PHP inside js ajax function diffuculties

PHP inside js ajax function diffuculties - javascript

I have button let's say like this:
<button id= $id onClick="popup($id)">button</button>
Now I am trying to use it with ajax function
<script>
function popup(id){
alert(" ");
document.write(" ");
}
</script>
All I want to do is to make this js function execute the php code.
I think puting it into alert of document.write would be good but this code looks like this below and i have big problems with quotations and everything. I have no idea honestly how to make it all work.
Could someone help me match it up please?
echo "<div id='element_to_pop_up'>
<a class='b-close'>zamknij</a>";
$messages = mysql_query('SELECT * FROM msg WHERE id = "'.$messageid.'" ');
$fetchmessage = mysql_fetch_assoc($messages);
echo $fetchmessage['message'];
echo "</div>";

PHP is processed on the server side. So you can not run PHP-code client side.
What I'd do is (here done in jQuery, you could do it with other frameworks as well) to set a data-id="" attribute on your button with the ID echoed from PHP. Then when you click the button, you call a PHP script via AJAX that does something. Like:
HTML:
<button class="clickme" data-id="<?=$row['id'];?>">button</button>
jQuery
$('.clickme').click(function(){
$.post( "test.php", { msgId: $(this).data('id') }).done(function( data ) {
alert("Data loaded: " + data);
});
});

You need something like this:
HTML
<button class="clickButton" data-id="<?php echo $id; ?>" id=""<?php echo $id; ?>">button</button>
jQuery
$(function() {
$('.clickButton').click(function() {
var id = $(this).data('id');
$.ajax({
type: "POST",
url: "ajax.php",
data: {id: id}
success: function(data) {
alert(data);
}
});
});
});
ajax.php
//Set up a mysqli connection here
echo "<div id='element_to_pop_up'><a class='b-close'>zamknij</a>";
$result = mysqli_query($link, "SELECT * FROM msg WHERE id = '" . mysqli_real_escape_string($link, $_POST["id"]) . "'");
$fetchmessage = mysqli_fetch_assoc($result);
echo $fetchmessage['message'];
echo "</div>";
Avoid to mysql injection
Use mysqli or PDO instead of mysql function since mysql functions are deprecated.

PHP runs on the server. JS runs in the browser. You will not run one from the other directly.
You can run PHP on the server that echoes a string that will be run as JS in the browser.
You can run JS that uses AJAX to call back to the server to run PHP on the server.
You cannot run PHP in the browser.
You cannot run browser JS on the server (for purposes of this discussion, ignoring all the ways you can).

Related

How can i use a Javascript variable inside my php code that is wrapped up into my script tag [duplicate]

I want to pass JavaScript variables to PHP using a hidden input in a form.
But I can't get the value of $_POST['hidden1'] into $salarieid. Is there something wrong?
Here is the code:
<script type="text/javascript">
// View what the user has chosen
function func_load3(name) {
var oForm = document.forms["myform"];
var oSelectBox = oForm.select3;
var iChoice = oSelectBox.selectedIndex;
//alert("You have chosen: " + oSelectBox.options[iChoice].text);
//document.write(oSelectBox.options[iChoice].text);
var sa = oSelectBox.options[iChoice].text;
document.getElementById("hidden1").value = sa;
}
</script>
<form name="myform" action="<?php echo $_SERVER['$PHP_SELF']; ?>" method="POST">
<input type="hidden" name="hidden1" id="hidden1" />
</form>
<?php
$salarieid = $_POST['hidden1'];
$query = "select * from salarie where salarieid = ".$salarieid;
echo $query;
$result = mysql_query($query);
?>
<table>
Code for displaying the query result.
</table>

You cannot pass variable values from the current page JavaScript code to the current page PHP code... PHP code runs at the server side, and it doesn't know anything about what is going on on the client side.
You need to pass variables to PHP code from the HTML form using another mechanism, such as submitting the form using the GET or POST methods.
<DOCTYPE html>
<html>
<head>
<title>My Test Form</title>
</head>
<body>
<form method="POST">
<p>Please, choose the salary id to proceed result:</p>
<p>
<label for="salarieids">SalarieID:</label>
<?php
$query = "SELECT * FROM salarie";
$result = mysql_query($query);
if ($result) :
?>
<select id="salarieids" name="salarieid">
<?php
while ($row = mysql_fetch_assoc($result)) {
echo '<option value="', $row['salaried'], '">', $row['salaried'], '</option>'; //between <option></option> tags you can output something more human-friendly (like $row['name'], if table "salaried" have one)
}
?>
</select>
<?php endif ?>
</p>
<p>
<input type="submit" value="Sumbit my choice"/>
</p>
</form>
<?php if isset($_POST['salaried']) : ?>
<?php
$query = "SELECT * FROM salarie WHERE salarieid = " . $_POST['salarieid'];
$result = mysql_query($query);
if ($result) :
?>
<table>
<?php
while ($row = mysql_fetch_assoc($result)) {
echo '<tr>';
echo '<td>', $row['salaried'], '</td><td>', $row['bla-bla-bla'], '</td>' ...; // and others
echo '</tr>';
}
?>
</table>
<?php endif?>
<?php endif ?>
</body>
</html>

Just save it in a cookie:
$(document).ready(function () {
createCookie("height", $(window).height(), "10");
});
function createCookie(name, value, days) {
var expires;
if (days) {
var date = new Date();
date.setTime(date.getTime() + (days * 24 * 60 * 60 * 1000));
expires = "; expires=" + date.toGMTString();
}
else {
expires = "";
}
document.cookie = escape(name) + "=" + escape(value) + expires + "; path=/";
}
And then read it with PHP:
<?PHP
$_COOKIE["height"];
?>
It's not a pretty solution, but it works.

There are several ways of passing variables from JavaScript to PHP (not the current page, of course).
You could:
Send the information in a form as stated here (will result in a page refresh)
Pass it in Ajax (several posts are on here about that) (without a page refresh)
Make an HTTP request via an XMLHttpRequest request (without a page refresh) like this:
if (window.XMLHttpRequest){
xmlhttp = new XMLHttpRequest();
}
else{
xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
}
var PageToSendTo = "nowitworks.php?";
var MyVariable = "variableData";
var VariablePlaceholder = "variableName=";
var UrlToSend = PageToSendTo + VariablePlaceholder + MyVariable;
xmlhttp.open("GET", UrlToSend, false);
xmlhttp.send();
I'm sure this could be made to look fancier and loop through all the variables and whatnot - but I've kept it basic as to make it easier to understand for the novices.

Here is the Working example: Get javascript variable value on the same page in php.
<script>
var p1 = "success";
</script>
<?php
echo "<script>document.writeln(p1);</script>";
?>

Here's how I did it (I needed to insert a local timezone into PHP:
<?php
ob_start();
?>
<script type="text/javascript">
var d = new Date();
document.write(d.getTimezoneOffset());
</script>
<?php
$offset = ob_get_clean();
print_r($offset);

When your page first loads the PHP code first runs and sets the complete layout of your webpage. After the page layout, it sets the JavaScript load up.
Now JavaScript directly interacts with DOM and can manipulate the layout but PHP can't - it needs to refresh the page. The only way is to refresh your page to and pass the parameters in the page URL so that you can get the data via PHP.
So, we use AJAX to get Javascript to interact with PHP without a page reload. AJAX can also be used as an API. One more thing if you have already declared the variable in PHP before the page loads then you can use it with your Javascript example.
<?php $myname= "syed ali";?>
<script>
var username = "<?php echo $myname;?>";
alert(username);
</script>
The above code is correct and it will work, but the code below is totally wrong and it will never work.
<script>
var username = "syed ali";
var <?php $myname;?> = username;
alert(myname);
</script>
Pass value from JavaScript to PHP via AJAX
This is the most secure way to do it, because HTML content can be edited via developer tools and the user can manipulate the data. So, it is better to use AJAX if you want security over that variable. If you are a newbie to AJAX, please learn AJAX it is very simple.
The best and most secure way to pass JavaScript variable into PHP is via AJAX
Simple AJAX example
var mydata = 55;
var myname = "syed ali";
var userdata = {'id':mydata,'name':myname};
$.ajax({
type: "POST",
url: "YOUR PHP URL HERE",
data:userdata,
success: function(data){
console.log(data);
}
});
PASS value from JavaScript to PHP via hidden fields
Otherwise, you can create a hidden HTML input inside your form. like
<input type="hidden" id="mydata">
then via jQuery or javaScript pass the value to the hidden field. like
<script>
var myvalue = 55;
$("#mydata").val(myvalue);
</script>
Now when you submit the form you can get the value in PHP.

I was trying to figure this out myself and then realized that the problem is that this is kind of a backwards way of looking at the situation. Rather than trying to pass things from JavaScript to php, maybe it's best to go the other way around, in most cases. PHP code executes on the server and creates the html code (and possibly java script as well). Then the browser loads the page and executes the html and java script.
It seems like the sensible way to approach situations like this is to use the PHP to create the JavaScript and the html you want and then to use the JavaScript in the page to do whatever PHP can't do. It seems like this would give you the benefits of both PHP and JavaScript in a fairly simple and straight forward way.
One thing I've done that gives the appearance of passing things to PHP from your page on the fly is using the html image tag to call on PHP code. Something like this:
<img src="pic.php">
The PHP code in pic.php would actually create html code before your web page was even loaded, but that html code is basically called upon on the fly. The php code here can be used to create a picture on your page, but it can have any commands you like besides that in it. Maybe it changes the contents of some files on your server, etc. The upside of this is that the php code can be executed from html and I assume JavaScript, but the down side is that the only output it can put on your page is an image. You also have the option of passing variables to the php code through parameters in the url. Page counters will use this technique in many cases.

PHP runs on the server before the page is sent to the user, JavaScript is run on the user's computer once it is received, so the PHP script has already executed.
If you want to pass a JavaScript value to a PHP script, you'd have to do an XMLHttpRequest to send the data back to the server.
Here's a previous question that you can follow for more information: Ajax Tutorial
Now if you just need to pass a form value to the server, you can also just do a normal form post, that does the same thing, but the whole page has to be refreshed.
<?php
if(isset($_POST))
{
print_r($_POST);
}
?>
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post">
<input type="text" name="data" value="1" />
<input type="submit" value="Submit" />
</form>
Clicking submit will submit the page, and print out the submitted data.

We can easily pass values even on same/ different pages using the cookies shown in the code as follows (In my case, I'm using it with facebook integration) -
function statusChangeCallback(response) {
console.log('statusChangeCallback');
if (response.status === 'connected') {
// Logged into your app and Facebook.
FB.api('/me?fields=id,first_name,last_name,email', function (result) {
document.cookie = "fbdata = " + result.id + "," + result.first_name + "," + result.last_name + "," + result.email;
console.log(document.cookie);
});
}
}
And I've accessed it (in any file) using -
<?php
if(isset($_COOKIE['fbdata'])) {
echo "welcome ".$_COOKIE['fbdata'];
}
?>

Your code has a few things wrong with it.
You define a JavaScript function, func_load3(), but do not call it.
Your function is defined in the wrong place. When it is defined in your page, the HTML objects it refers to have not yet been loaded. Most JavaScript code checks whether the document is fully loaded before executing, or you can just move your code past the elements it refers to in the page.
Your form has no means to submit it. It needs a submit button.
You do not check whether your form has been submitted.
It is possible to set a JavaScript variable in a hidden variable in a form, then submit it, and read the value back in PHP. Here is a simple example that shows this:
<?php
if (isset($_POST['hidden1'])) {
echo "You submitted {$_POST['hidden1']}";
die;
}
echo <<<HTML
<form name="myform" action="{$_SERVER['PHP_SELF']}" method="post" id="myform">
<input type="submit" name="submit" value="Test this mess!" />
<input type="hidden" name="hidden1" id="hidden1" />
</form>
<script type="text/javascript">
document.getElementById("hidden1").value = "This is an example";
</script>
HTML;
?>

You can use JQuery Ajax and POST method:
var obj;
$(document).ready(function(){
$("#button1").click(function(){
var username=$("#username").val();
var password=$("#password").val();
$.ajax({
url: "addperson.php",
type: "POST",
async: false,
data: {
username: username,
password: password
}
})
.done (function(data, textStatus, jqXHR) {
obj = JSON.parse(data);
})
.fail (function(jqXHR, textStatus, errorThrown) {
})
.always (function(jqXHROrData, textStatus, jqXHROrErrorThrown) {
});
});
});
To take a response back from the php script JSON parse the the respone in .done() method.
Here is the php script you can modify to your needs:
<?php
$username1 = isset($_POST["username"]) ? $_POST["username"] : '';
$password1 = isset($_POST["password"]) ? $_POST["password"] : '';
$servername = "xxxxx";
$username = "xxxxx";
$password = "xxxxx";
$dbname = "xxxxx";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "INSERT INTO user (username, password)
VALUES ('$username1', '$password1' )";
;
if ($conn->query($sql) === TRUE) {
echo json_encode(array('success' => 1));
} else{
echo json_encode(array('success' => 0));
}
$conn->close();
?>

Is your function, which sets the hidden form value, being called? It is not in this example. You should have no problem modifying a hidden value before posting the form back to the server.

May be you could use jquery serialize() method so that everything will be at one go.
var data=$('#myForm').serialize();
//this way you could get the hidden value as well in the server side.

This obviously solution was not mentioned earlier. You can also use cookies to pass data from the browser back to the server.
Just set a cookie with the data you want to pass to PHP using javascript in the browser.
Then, simply read this cookie on the PHP side.

We cannot pass JavaScript variable values to the PHP code directly... PHP code runs at the server side, and it doesn't know anything about what is going on on the client side.
So it's better to use the AJAX to parse the JavaScript value into the php Code.
Or alternatively we can make this done with the help of COOKIES in our code.
Thanks & Cheers.

Use the + sign to concatenate your javascript variable into your php function call.
<script>
var JSvar = "success";
var JSnewVar = "<?=myphpFunction('" + JSvar + "');?>";
</script>`
Notice the = sign is there twice.

Passing a javascript variable to a php variable on the same page using ajax [duplicate]

I want to pass JavaScript variables to PHP using a hidden input in a form.
But I can't get the value of $_POST['hidden1'] into $salarieid. Is there something wrong?
Here is the code:
<script type="text/javascript">
// View what the user has chosen
function func_load3(name) {
var oForm = document.forms["myform"];
var oSelectBox = oForm.select3;
var iChoice = oSelectBox.selectedIndex;
//alert("You have chosen: " + oSelectBox.options[iChoice].text);
//document.write(oSelectBox.options[iChoice].text);
var sa = oSelectBox.options[iChoice].text;
document.getElementById("hidden1").value = sa;
}
</script>
<form name="myform" action="<?php echo $_SERVER['$PHP_SELF']; ?>" method="POST">
<input type="hidden" name="hidden1" id="hidden1" />
</form>
<?php
$salarieid = $_POST['hidden1'];
$query = "select * from salarie where salarieid = ".$salarieid;
echo $query;
$result = mysql_query($query);
?>
<table>
Code for displaying the query result.
</table>

You cannot pass variable values from the current page JavaScript code to the current page PHP code... PHP code runs at the server side, and it doesn't know anything about what is going on on the client side.
You need to pass variables to PHP code from the HTML form using another mechanism, such as submitting the form using the GET or POST methods.
<DOCTYPE html>
<html>
<head>
<title>My Test Form</title>
</head>
<body>
<form method="POST">
<p>Please, choose the salary id to proceed result:</p>
<p>
<label for="salarieids">SalarieID:</label>
<?php
$query = "SELECT * FROM salarie";
$result = mysql_query($query);
if ($result) :
?>
<select id="salarieids" name="salarieid">
<?php
while ($row = mysql_fetch_assoc($result)) {
echo '<option value="', $row['salaried'], '">', $row['salaried'], '</option>'; //between <option></option> tags you can output something more human-friendly (like $row['name'], if table "salaried" have one)
}
?>
</select>
<?php endif ?>
</p>
<p>
<input type="submit" value="Sumbit my choice"/>
</p>
</form>
<?php if isset($_POST['salaried']) : ?>
<?php
$query = "SELECT * FROM salarie WHERE salarieid = " . $_POST['salarieid'];
$result = mysql_query($query);
if ($result) :
?>
<table>
<?php
while ($row = mysql_fetch_assoc($result)) {
echo '<tr>';
echo '<td>', $row['salaried'], '</td><td>', $row['bla-bla-bla'], '</td>' ...; // and others
echo '</tr>';
}
?>
</table>
<?php endif?>
<?php endif ?>
</body>
</html>

Just save it in a cookie:
$(document).ready(function () {
createCookie("height", $(window).height(), "10");
});
function createCookie(name, value, days) {
var expires;
if (days) {
var date = new Date();
date.setTime(date.getTime() + (days * 24 * 60 * 60 * 1000));
expires = "; expires=" + date.toGMTString();
}
else {
expires = "";
}
document.cookie = escape(name) + "=" + escape(value) + expires + "; path=/";
}
And then read it with PHP:
<?PHP
$_COOKIE["height"];
?>
It's not a pretty solution, but it works.

There are several ways of passing variables from JavaScript to PHP (not the current page, of course).
You could:
Send the information in a form as stated here (will result in a page refresh)
Pass it in Ajax (several posts are on here about that) (without a page refresh)
Make an HTTP request via an XMLHttpRequest request (without a page refresh) like this:
if (window.XMLHttpRequest){
xmlhttp = new XMLHttpRequest();
}
else{
xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
}
var PageToSendTo = "nowitworks.php?";
var MyVariable = "variableData";
var VariablePlaceholder = "variableName=";
var UrlToSend = PageToSendTo + VariablePlaceholder + MyVariable;
xmlhttp.open("GET", UrlToSend, false);
xmlhttp.send();
I'm sure this could be made to look fancier and loop through all the variables and whatnot - but I've kept it basic as to make it easier to understand for the novices.

Here is the Working example: Get javascript variable value on the same page in php.
<script>
var p1 = "success";
</script>
<?php
echo "<script>document.writeln(p1);</script>";
?>

Here's how I did it (I needed to insert a local timezone into PHP:
<?php
ob_start();
?>
<script type="text/javascript">
var d = new Date();
document.write(d.getTimezoneOffset());
</script>
<?php
$offset = ob_get_clean();
print_r($offset);

When your page first loads the PHP code first runs and sets the complete layout of your webpage. After the page layout, it sets the JavaScript load up.
Now JavaScript directly interacts with DOM and can manipulate the layout but PHP can't - it needs to refresh the page. The only way is to refresh your page to and pass the parameters in the page URL so that you can get the data via PHP.
So, we use AJAX to get Javascript to interact with PHP without a page reload. AJAX can also be used as an API. One more thing if you have already declared the variable in PHP before the page loads then you can use it with your Javascript example.
<?php $myname= "syed ali";?>
<script>
var username = "<?php echo $myname;?>";
alert(username);
</script>
The above code is correct and it will work, but the code below is totally wrong and it will never work.
<script>
var username = "syed ali";
var <?php $myname;?> = username;
alert(myname);
</script>
Pass value from JavaScript to PHP via AJAX
This is the most secure way to do it, because HTML content can be edited via developer tools and the user can manipulate the data. So, it is better to use AJAX if you want security over that variable. If you are a newbie to AJAX, please learn AJAX it is very simple.
The best and most secure way to pass JavaScript variable into PHP is via AJAX
Simple AJAX example
var mydata = 55;
var myname = "syed ali";
var userdata = {'id':mydata,'name':myname};
$.ajax({
type: "POST",
url: "YOUR PHP URL HERE",
data:userdata,
success: function(data){
console.log(data);
}
});
PASS value from JavaScript to PHP via hidden fields
Otherwise, you can create a hidden HTML input inside your form. like
<input type="hidden" id="mydata">
then via jQuery or javaScript pass the value to the hidden field. like
<script>
var myvalue = 55;
$("#mydata").val(myvalue);
</script>
Now when you submit the form you can get the value in PHP.

PHP runs on the server before the page is sent to the user, JavaScript is run on the user's computer once it is received, so the PHP script has already executed.
If you want to pass a JavaScript value to a PHP script, you'd have to do an XMLHttpRequest to send the data back to the server.
Here's a previous question that you can follow for more information: Ajax Tutorial
Now if you just need to pass a form value to the server, you can also just do a normal form post, that does the same thing, but the whole page has to be refreshed.
<?php
if(isset($_POST))
{
print_r($_POST);
}
?>
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post">
<input type="text" name="data" value="1" />
<input type="submit" value="Submit" />
</form>
Clicking submit will submit the page, and print out the submitted data.

We can easily pass values even on same/ different pages using the cookies shown in the code as follows (In my case, I'm using it with facebook integration) -
function statusChangeCallback(response) {
console.log('statusChangeCallback');
if (response.status === 'connected') {
// Logged into your app and Facebook.
FB.api('/me?fields=id,first_name,last_name,email', function (result) {
document.cookie = "fbdata = " + result.id + "," + result.first_name + "," + result.last_name + "," + result.email;
console.log(document.cookie);
});
}
}
And I've accessed it (in any file) using -
<?php
if(isset($_COOKIE['fbdata'])) {
echo "welcome ".$_COOKIE['fbdata'];
}
?>

Your code has a few things wrong with it.
You define a JavaScript function, func_load3(), but do not call it.
Your function is defined in the wrong place. When it is defined in your page, the HTML objects it refers to have not yet been loaded. Most JavaScript code checks whether the document is fully loaded before executing, or you can just move your code past the elements it refers to in the page.
Your form has no means to submit it. It needs a submit button.
You do not check whether your form has been submitted.
It is possible to set a JavaScript variable in a hidden variable in a form, then submit it, and read the value back in PHP. Here is a simple example that shows this:
<?php
if (isset($_POST['hidden1'])) {
echo "You submitted {$_POST['hidden1']}";
die;
}
echo <<<HTML
<form name="myform" action="{$_SERVER['PHP_SELF']}" method="post" id="myform">
<input type="submit" name="submit" value="Test this mess!" />
<input type="hidden" name="hidden1" id="hidden1" />
</form>
<script type="text/javascript">
document.getElementById("hidden1").value = "This is an example";
</script>
HTML;
?>

You can use JQuery Ajax and POST method:
var obj;
$(document).ready(function(){
$("#button1").click(function(){
var username=$("#username").val();
var password=$("#password").val();
$.ajax({
url: "addperson.php",
type: "POST",
async: false,
data: {
username: username,
password: password
}
})
.done (function(data, textStatus, jqXHR) {
obj = JSON.parse(data);
})
.fail (function(jqXHR, textStatus, errorThrown) {
})
.always (function(jqXHROrData, textStatus, jqXHROrErrorThrown) {
});
});
});
To take a response back from the php script JSON parse the the respone in .done() method.
Here is the php script you can modify to your needs:
<?php
$username1 = isset($_POST["username"]) ? $_POST["username"] : '';
$password1 = isset($_POST["password"]) ? $_POST["password"] : '';
$servername = "xxxxx";
$username = "xxxxx";
$password = "xxxxx";
$dbname = "xxxxx";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "INSERT INTO user (username, password)
VALUES ('$username1', '$password1' )";
;
if ($conn->query($sql) === TRUE) {
echo json_encode(array('success' => 1));
} else{
echo json_encode(array('success' => 0));
}
$conn->close();
?>

Is your function, which sets the hidden form value, being called? It is not in this example. You should have no problem modifying a hidden value before posting the form back to the server.

May be you could use jquery serialize() method so that everything will be at one go.
var data=$('#myForm').serialize();
//this way you could get the hidden value as well in the server side.

This obviously solution was not mentioned earlier. You can also use cookies to pass data from the browser back to the server.
Just set a cookie with the data you want to pass to PHP using javascript in the browser.
Then, simply read this cookie on the PHP side.

We cannot pass JavaScript variable values to the PHP code directly... PHP code runs at the server side, and it doesn't know anything about what is going on on the client side.
So it's better to use the AJAX to parse the JavaScript value into the php Code.
Or alternatively we can make this done with the help of COOKIES in our code.
Thanks & Cheers.

Use the + sign to concatenate your javascript variable into your php function call.
<script>
var JSvar = "success";
var JSnewVar = "<?=myphpFunction('" + JSvar + "');?>";
</script>`
Notice the = sign is there twice.

How to send or pass a javascript variable to a php file? [duplicate]

I want to pass JavaScript variables to PHP using a hidden input in a form.
But I can't get the value of $_POST['hidden1'] into $salarieid. Is there something wrong?
Here is the code:
<script type="text/javascript">
// View what the user has chosen
function func_load3(name) {
var oForm = document.forms["myform"];
var oSelectBox = oForm.select3;
var iChoice = oSelectBox.selectedIndex;
//alert("You have chosen: " + oSelectBox.options[iChoice].text);
//document.write(oSelectBox.options[iChoice].text);
var sa = oSelectBox.options[iChoice].text;
document.getElementById("hidden1").value = sa;
}
</script>
<form name="myform" action="<?php echo $_SERVER['$PHP_SELF']; ?>" method="POST">
<input type="hidden" name="hidden1" id="hidden1" />
</form>
<?php
$salarieid = $_POST['hidden1'];
$query = "select * from salarie where salarieid = ".$salarieid;
echo $query;
$result = mysql_query($query);
?>
<table>
Code for displaying the query result.
</table>

You cannot pass variable values from the current page JavaScript code to the current page PHP code... PHP code runs at the server side, and it doesn't know anything about what is going on on the client side.
You need to pass variables to PHP code from the HTML form using another mechanism, such as submitting the form using the GET or POST methods.
<DOCTYPE html>
<html>
<head>
<title>My Test Form</title>
</head>
<body>
<form method="POST">
<p>Please, choose the salary id to proceed result:</p>
<p>
<label for="salarieids">SalarieID:</label>
<?php
$query = "SELECT * FROM salarie";
$result = mysql_query($query);
if ($result) :
?>
<select id="salarieids" name="salarieid">
<?php
while ($row = mysql_fetch_assoc($result)) {
echo '<option value="', $row['salaried'], '">', $row['salaried'], '</option>'; //between <option></option> tags you can output something more human-friendly (like $row['name'], if table "salaried" have one)
}
?>
</select>
<?php endif ?>
</p>
<p>
<input type="submit" value="Sumbit my choice"/>
</p>
</form>
<?php if isset($_POST['salaried']) : ?>
<?php
$query = "SELECT * FROM salarie WHERE salarieid = " . $_POST['salarieid'];
$result = mysql_query($query);
if ($result) :
?>
<table>
<?php
while ($row = mysql_fetch_assoc($result)) {
echo '<tr>';
echo '<td>', $row['salaried'], '</td><td>', $row['bla-bla-bla'], '</td>' ...; // and others
echo '</tr>';
}
?>
</table>
<?php endif?>
<?php endif ?>
</body>
</html>

Just save it in a cookie:
$(document).ready(function () {
createCookie("height", $(window).height(), "10");
});
function createCookie(name, value, days) {
var expires;
if (days) {
var date = new Date();
date.setTime(date.getTime() + (days * 24 * 60 * 60 * 1000));
expires = "; expires=" + date.toGMTString();
}
else {
expires = "";
}
document.cookie = escape(name) + "=" + escape(value) + expires + "; path=/";
}
And then read it with PHP:
<?PHP
$_COOKIE["height"];
?>
It's not a pretty solution, but it works.

There are several ways of passing variables from JavaScript to PHP (not the current page, of course).
You could:
Send the information in a form as stated here (will result in a page refresh)
Pass it in Ajax (several posts are on here about that) (without a page refresh)
Make an HTTP request via an XMLHttpRequest request (without a page refresh) like this:
if (window.XMLHttpRequest){
xmlhttp = new XMLHttpRequest();
}
else{
xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
}
var PageToSendTo = "nowitworks.php?";
var MyVariable = "variableData";
var VariablePlaceholder = "variableName=";
var UrlToSend = PageToSendTo + VariablePlaceholder + MyVariable;
xmlhttp.open("GET", UrlToSend, false);
xmlhttp.send();
I'm sure this could be made to look fancier and loop through all the variables and whatnot - but I've kept it basic as to make it easier to understand for the novices.

Here is the Working example: Get javascript variable value on the same page in php.
<script>
var p1 = "success";
</script>
<?php
echo "<script>document.writeln(p1);</script>";
?>

Here's how I did it (I needed to insert a local timezone into PHP:
<?php
ob_start();
?>
<script type="text/javascript">
var d = new Date();
document.write(d.getTimezoneOffset());
</script>
<?php
$offset = ob_get_clean();
print_r($offset);

When your page first loads the PHP code first runs and sets the complete layout of your webpage. After the page layout, it sets the JavaScript load up.
Now JavaScript directly interacts with DOM and can manipulate the layout but PHP can't - it needs to refresh the page. The only way is to refresh your page to and pass the parameters in the page URL so that you can get the data via PHP.
So, we use AJAX to get Javascript to interact with PHP without a page reload. AJAX can also be used as an API. One more thing if you have already declared the variable in PHP before the page loads then you can use it with your Javascript example.
<?php $myname= "syed ali";?>
<script>
var username = "<?php echo $myname;?>";
alert(username);
</script>
The above code is correct and it will work, but the code below is totally wrong and it will never work.
<script>
var username = "syed ali";
var <?php $myname;?> = username;
alert(myname);
</script>
Pass value from JavaScript to PHP via AJAX
This is the most secure way to do it, because HTML content can be edited via developer tools and the user can manipulate the data. So, it is better to use AJAX if you want security over that variable. If you are a newbie to AJAX, please learn AJAX it is very simple.
The best and most secure way to pass JavaScript variable into PHP is via AJAX
Simple AJAX example
var mydata = 55;
var myname = "syed ali";
var userdata = {'id':mydata,'name':myname};
$.ajax({
type: "POST",
url: "YOUR PHP URL HERE",
data:userdata,
success: function(data){
console.log(data);
}
});
PASS value from JavaScript to PHP via hidden fields
Otherwise, you can create a hidden HTML input inside your form. like
<input type="hidden" id="mydata">
then via jQuery or javaScript pass the value to the hidden field. like
<script>
var myvalue = 55;
$("#mydata").val(myvalue);
</script>
Now when you submit the form you can get the value in PHP.

PHP runs on the server before the page is sent to the user, JavaScript is run on the user's computer once it is received, so the PHP script has already executed.
If you want to pass a JavaScript value to a PHP script, you'd have to do an XMLHttpRequest to send the data back to the server.
Here's a previous question that you can follow for more information: Ajax Tutorial
Now if you just need to pass a form value to the server, you can also just do a normal form post, that does the same thing, but the whole page has to be refreshed.
<?php
if(isset($_POST))
{
print_r($_POST);
}
?>
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post">
<input type="text" name="data" value="1" />
<input type="submit" value="Submit" />
</form>
Clicking submit will submit the page, and print out the submitted data.

We can easily pass values even on same/ different pages using the cookies shown in the code as follows (In my case, I'm using it with facebook integration) -
function statusChangeCallback(response) {
console.log('statusChangeCallback');
if (response.status === 'connected') {
// Logged into your app and Facebook.
FB.api('/me?fields=id,first_name,last_name,email', function (result) {
document.cookie = "fbdata = " + result.id + "," + result.first_name + "," + result.last_name + "," + result.email;
console.log(document.cookie);
});
}
}
And I've accessed it (in any file) using -
<?php
if(isset($_COOKIE['fbdata'])) {
echo "welcome ".$_COOKIE['fbdata'];
}
?>

Your code has a few things wrong with it.
You define a JavaScript function, func_load3(), but do not call it.
Your function is defined in the wrong place. When it is defined in your page, the HTML objects it refers to have not yet been loaded. Most JavaScript code checks whether the document is fully loaded before executing, or you can just move your code past the elements it refers to in the page.
Your form has no means to submit it. It needs a submit button.
You do not check whether your form has been submitted.
It is possible to set a JavaScript variable in a hidden variable in a form, then submit it, and read the value back in PHP. Here is a simple example that shows this:
<?php
if (isset($_POST['hidden1'])) {
echo "You submitted {$_POST['hidden1']}";
die;
}
echo <<<HTML
<form name="myform" action="{$_SERVER['PHP_SELF']}" method="post" id="myform">
<input type="submit" name="submit" value="Test this mess!" />
<input type="hidden" name="hidden1" id="hidden1" />
</form>
<script type="text/javascript">
document.getElementById("hidden1").value = "This is an example";
</script>
HTML;
?>

You can use JQuery Ajax and POST method:
var obj;
$(document).ready(function(){
$("#button1").click(function(){
var username=$("#username").val();
var password=$("#password").val();
$.ajax({
url: "addperson.php",
type: "POST",
async: false,
data: {
username: username,
password: password
}
})
.done (function(data, textStatus, jqXHR) {
obj = JSON.parse(data);
})
.fail (function(jqXHR, textStatus, errorThrown) {
})
.always (function(jqXHROrData, textStatus, jqXHROrErrorThrown) {
});
});
});
To take a response back from the php script JSON parse the the respone in .done() method.
Here is the php script you can modify to your needs:
<?php
$username1 = isset($_POST["username"]) ? $_POST["username"] : '';
$password1 = isset($_POST["password"]) ? $_POST["password"] : '';
$servername = "xxxxx";
$username = "xxxxx";
$password = "xxxxx";
$dbname = "xxxxx";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "INSERT INTO user (username, password)
VALUES ('$username1', '$password1' )";
;
if ($conn->query($sql) === TRUE) {
echo json_encode(array('success' => 1));
} else{
echo json_encode(array('success' => 0));
}
$conn->close();
?>

Is your function, which sets the hidden form value, being called? It is not in this example. You should have no problem modifying a hidden value before posting the form back to the server.

May be you could use jquery serialize() method so that everything will be at one go.
var data=$('#myForm').serialize();
//this way you could get the hidden value as well in the server side.

This obviously solution was not mentioned earlier. You can also use cookies to pass data from the browser back to the server.
Just set a cookie with the data you want to pass to PHP using javascript in the browser.
Then, simply read this cookie on the PHP side.

We cannot pass JavaScript variable values to the PHP code directly... PHP code runs at the server side, and it doesn't know anything about what is going on on the client side.
So it's better to use the AJAX to parse the JavaScript value into the php Code.
Or alternatively we can make this done with the help of COOKIES in our code.
Thanks & Cheers.

Use the + sign to concatenate your javascript variable into your php function call.
<script>
var JSvar = "success";
var JSnewVar = "<?=myphpFunction('" + JSvar + "');?>";
</script>`
Notice the = sign is there twice.

getting POST data through AJAX

I have a form that will go through AJAX to set a session variable and back to the original page.
My problem is that I cannot set the session variable, thus, it is not working.
my main PHP page
<?php
if ($_SESSION['adminFunction'] == 'adminfunction'){
echo "<form id='userOptionRequest' action='request.php'>";
echo "<button name='adminFuncUserOption' onclick='adminFuncOption(1)'>User Option</button> <br /> <br />";
echo "<button name='adminFuncUserOption' onclick='adminFuncOption(2)'>Subject Option</button> <br /> <br />";
echo "<button name='adminFuncUserOption' onclick='adminFuncOption(3)'>Test Option</button> <br /> <br />";
echo "</form>";
}
if ($_SESSION['adminFunction'] == 'addusers'){
echo "hello world";
}
?>
my JS file
function adminFuncOption(option){
if (option == 1){
var userOptionRequest = new XMLHttpRequest();
userOptionRequest.open('POST','request.php',false);
xmlhttp.setRequestHeader('Content-type','application/x-www-form-urlencoded');
userOptionRequest.send('adminFuncUserOption=' + option);
return(1);
} }
the request.php
<?php
session_start();
$option = $_POST['option'];
if ($option != NULL){
if ($option == 1){
$_SESSION['adminFunction'] = 'addusers';
header('Location: http://rsc_naga_isd/admin');
}
}
var_dump($_POST);
var_dump($_SESSION);
var_dump($_GET);
?>
when I go back to my main PHP page, the session should now be addusers and should display 'hello world'
I have research but still had a hard time understanding AJAX or PHP
Submitting HTML form using Jquery AJAX
http://www.ajax-tutor.com/post-data-server.html
http://tutorialzine.com/2009/09/simple-ajax-website-jquery/
Oh, am still not familiar with jquery and still new to javascript, but would prefer javascript before diving in to jquery.

Hi jquery helps you to do more with less for example an AJAX call using jquery is like this:
$.ajax({
url: url,
type: 'POST',
data: data,
beforeSend: funtion(){
//DO Something before send like print a Loading text
},
success: function(data){
//DO SOMETHING IF DATA IS RETURNED
},
error: function(data){
//DO SOMETHING ON ERROR
}
});
Then you can check on you php scipt for the values sent on data for example:
if you send this data value:
data: {"adminFunction" : "mortal user not worthy" }
on your php script just do something like this:
switch($_POST['adminFunction'])
{
case 'superAdmin':
//....
break;
default:
//DO Something
break;
}
More info here:
AJAX JQUERY

In javascript you send:
userOptionRequest.send('adminFuncUserOption=' + option);
but in php:
$option = $_POST['option'];
it should be
$option = $_POST['adminFuncUserOption'];
I know you want javascript, but jquery so much easier ;)
$.post("request.php", { adminFuncUserOption: option }, function(response) {
console.log(response);
});
that's it then just check for $_POST['adminFuncUserOption']

The server side of the code (PHP) executes prior to the javascript side of the code. The server creates HTML (with javascript) to the client browser. The client browser then executes the code and javascript.
While your Session variable was updated, the requesting page was already created. You are going to need to reload the page that made the request in order to recognize the change to the session variable.
You could have the AJAX pass back non-sensitive data that you can process with javascript. If this is secure information that you don't want to be hacked, then I recommend avoiding AJAX until you get a better handle of it.

How to call AJAX in a WordPress plugin? [duplicate]

This question already has an answer here:
How do I use Ajax in a Wordpress plugin?
(1 answer)
Closed 8 years ago.
I want to fetch MySQL table (custom table) data to my admin panel via AJAX. When user click the button get data without page refresh.
I already created it on PHP but where to add AJAX file in WordPress?
This is my code. Note I am trying to make plugin When user click fetchdata button call the AJAX for result.
I added JS file in my plugin
I called js from main plugin file
function school_register_head() {
$siteurl = get_option('siteurl');
$url2 = $siteurl . '/wp-content/plugins/' . basename(dirname(__FILE__)) . '/jquery-1.3.2.js';
$url3 = $siteurl . '/wp-content/plugins/' . basename(dirname(__FILE__)) . '/script.js';
echo "<script src=$url2 type=text/javascript></script>\n";
echo "<script src=$url3 type=text/javascript></script>\n";
}
add_action('admin_head', 'school_register_head');
script.js file
$(document).ready(function() {
$("#display").click(function() {
$.ajax({ //create an ajax request to load_page.php
type: "GET",
url: "display.php",
dataType: "html", //expect html to be returned
success: function(response){
$("#responsecontainer").html(response);
//alert(response);
}
});
});
});
display.php file
<?php
global $wpdb;
$rows = $wpdb->get_results("SELECT postid from `wp_school_post`");
echo "<table class='wp-list-table widefat fixed'>";
echo "<tr><th>ID</th><tr>";
foreach ($rows as $row ){
echo "<tr>";
echo "<td>$row->postid</td>";
echo "</tr>";}
echo "</table>";
?>
html button
<input type="button" id="display" class="button" value="Fetch All Data" onClick="fetch_data();" />
<div id="responsecontainer" align="center">

If "where to put the AJAX file" means how to include it to the admin page: The easiest way would be to use admin_enqueue_scripts(), for example, if you're implementing it in a theme:
<?php
function load_ajax_script() {
wp_enqueue_script("ajax-script", get_template_directory() . "/scripts/ajax-script.js");
}
add_action("admin_enqueue_scripts", "load_ajax_script");
?>
Add this code to your functions.php within the active theme directory.
Next step would be the jQuery script. Open the .js script file used with wp_enqueue_script() and add the following:
jQuery(document).ready(function() {
jQuery("#fetch-data").click(function() {
var script_url = "http://www.example.com/wp-content/themes/my-theme/display.php";
var response_container = jQuery("#responsecontainer");
jQuery.ajax({
type: "GET",
url:script_url,
success:function(response) {
response_container.html(response);
},
error:function() {
alert("There was an error while fetching the data.");
}
});
});
});
Note that you have to use a URL when calling the .php file, not the absolute path on the server as you would when using PHP's include(). Depending on where you put the script, use either $(...) or jQuery(...).
Now we have to insert the button that is calling AJAX and of course the container where the returned content is being inserted. Since you didn't say where it is being displayed, I just put it in the top of the admin area for now:
function display_ajax_button() {
?>
<input type="button" value="Fetch data" id="fetch-data" />
<div id="fetched-data-content"></div>
<?php
}
add_action("admin_init", "display_ajax_button");
Finally, we create the PHP script itself. I've just copied your code and saved it in the same directory as defined in the jQuery.ajax() URL parameter:
<?php
global $wpdb;
$rows = $wpdb->get_results("SELECT postid from `wp_school_post`");
echo "<table class='wp-list-table widefat fixed'>";
echo "<tr><th>ID</th><tr>";
foreach ($rows as $row ){
echo "<tr>";
echo "<td>$row->postid</td>";
echo "</tr>";}
echo "</table>";
?>
That should be the simpliest approach. Open the admin page and give it a try. Let me know if it worked or not.

Develop Reference

JavaScript is the programming language of the Web.