Related
I posted this earlier but it incorrectly got marked as a duplicate of Can scripts be inserted with innerHTML?, which isn't my problem. I'm not trying to run any JavaScript through using innerHTML, I'm trying to run JavaScript that is located in a PHP file called through Ajax/XmlHttp. I am using innerHTML in that JS, but I'm not writing more JS within that. So it's not a duplicate of that question and I'm trying it again now.
Not sure what's going on here. I'll explain the organization of my files first and then get into the code.
The application- Allows the user to select different attributes (i.e. year or name) and view pictures of matching results from a database.
Files- I have a gallery.php file that has a series of HTML form elements acting as selectors to get filtered results from a database. Whenever a selector is set or changed, the file sends a new request to a get.php file that uses Ajax to refresh the results without loading a new page. All that works great. My next task is to implement a modal section where I can click on an image and view a bigger version which I plan to do with JavaScript and CSS, but my intermediate goal is to just change some text at the bottom of the results from get.php again using JavaScript, just as a first step. But I can't seem to get any JavaScript written in get.php to fire.
Code-
This is gallery.php:
<?php
include_once("./../php/navbar.php");
?>
<html>
<head>
<link href="/css/siteTheme.css" rel="stylesheet">
<style>
.attributes span {
margin-right: 1rem;
}
</style>
<script>
function changeParams() {
if (window.XMLHttpRequest) {
xmlhttp = new XMLHttpRequest();
} else {
xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange = function() {
if (this.readyState == 4 && this.status == 200) {
document.getElementById("test").innerHTML = this.responseText;
}
};
year = document.getElementById("years_select").value;
nameFirst= document.getElementById("nameFirst_select").value;
/* Ton of other getElementById statements here that I'm excluding to keep things shorter */
url = "get.php?year="+year+......;
xmlhttp.open("GET", url, true);
xmlhttp.send();
} /* End function */
</script>
</head>
<body>
<div class="content">
<h1>Gallery</h1>
<form>Details:
<!-- -------------------- Year -------------------- -->
<select onchange="changeParams()" name="years" id="years_select">
<option value="All">Year</option>
<?php
include("/var/www/admin.php");
$conn = mysqli_connect($dbServername, $publicdbUsername, $publicdbPass, $dbName);
if (!$conn) {
die('Could not connect: ' . mysqli_error($conn));
}
$sql = "select year from db group by year order by year desc";
$result = mysqli_query($conn, $sql);
while ($row = mysqli_fetch_array($result)) {
echo "<option value=" . $row['year'] . ">" . $row['year'] . "</option>";
}
mysqli_close($conn);
?>
</select>
<!-- A bunch of other selectors are created here following the same pattern as above of getting results from a DB -->
<!-- -------------------- Searchbar -------------------- -->
<input type="text" size="50" onkeyup="changeParams()" name="search" id="search" placeholder="Search"></input>
</form>
<div id="test">Choose some filters to see some results!</div>
</form>
</div> <!-- Ends content div -->
</body>
</html>
This is get.php:
<html>
<head>
<style>
/* Bunch of CSS that's not relevant here */
</style>
<script type="text/javascript">
console.log(".....");
var parts = document.querySelectorAll("div.imgContainer");
console.log("Found something:", parts);
parts.addEventListener("click", function(){
document.getElementById("anID").innerHTML = "Test...";
});
</script>
</head>
<body>
<?php
include("/var/www/admin.php");
$year = $_GET['year'];
//Bunch of other variables set here following the same logic, getting data from gallery.php
$conn = mysqli_connect($dbServername, $publicdbUsername, $publicdbPass, $dbName);
if (!$conn) {
die('Could not connect: ' . mysqli_error($conn));
echo '$conn';
}
$sql = 'select * db';
/* ---------- Creating SQL statement ---------- */
$clauses = 0;
if ($year != "All") {
if ($clauses == 0) {
$sql = $sql . ' where year = "' . $year . '" and';
$clauses = $clauses + 1;
} else {
$sql = $sql . ' year = "' . $year . '" and';
}
} /* Bunch of other if statements to get set information and add to sql statement as such */
// Need to chop of the last ' and' from the sql statement
$sql = substr($sql, 0, -4);
$sql = $sql . ' order by year desc';
$result = mysqli_query($conn, $sql);
$num_results = mysqli_num_rows($result);
if ($num_results == 0 or $clauses == 0) {
echo "<p>No matches to your query. Try refining your search terms to get some results.</p>";
} else {
echo "<p>" . $num_results . " results matched your query.</p>";
echo "<div class=results>";
//echo "<div>";
echo '<script type="text/javascript">
function modalFunction() {
document.getElementById("anID").innerHTML = "test";
}
</script>';
while ($row = mysqli_fetch_array($result)) {
$pic = $row['pathToPic'];
$wwwImg = substr($pic, 13);
//echo "<span id=aCard><img src=" . $wwwImg . " height ='250px'>";
//echo "<span class=text>" . $row['fullCardInfo'] . "</span></span>";
echo "<div class=fullContainer><div class='imgContainer'><img class=image src=" . $wwwImg ."></div><p class=text>" . $row['fullInfo'] . "</p></div>";
} // End while of results
echo "</div>";// End results div//</div>";
//echo '<div class="modal"><p id="anID"></p></div>';
} // End else of "if results"
mysqli_close($conn);
?>
<script>
</script>
<div>
<p id="anID">This in a div</p>
</div>
<!--<span>
<p id="anID">This in a span</p>
</span>-->
</body>
</html>
Sorry if that was messy, I chopped out a bunch of stuff that just gets/selects/filters some data. All that works and all variables that are left in there are set in my full code.
But the issue I'm having is writing JavaScript in get.php to change the text in the <p id="anID"> tags. I've tried JS in a few different areas of get.php, from in the <head> tags, to echoing it in <script> tags in the PHP, to pure <script> tags after the PHP statements are done (I think I left them in a few different places).
Problem is that nothing I do works to change the text in the <p> tags I references. Additionally, the console.log statements in the header of get.php don't seem to get called either. They don't fire no matter where I place them in get.php. Console.log works fine in gallery.php so it's not some issue with my browser or anything.
Long term, I will be adding JS query selectors to bring up a bigger image when an image is clicked on, but for now I'm just trying to make ANY JavaScript work in get.php and I'm struggling to do so. I don't see how this is a duplicate of the suggested repeat as I'm not trying to run JavaScript through innerHTML here.
Scratch that, it actually is trying to pass JavaScript through innerHTML as the information from get.php comes from this section of gallery.php:
xmlhttp.onreadystatechange = function() {
if (this.readyState == 4 && this.status == 200) {
document.getElementById("test").innerHTML = this.responseText;
console.log("Response: ", this.responseText);
}
So it appears this is a duplicate of the question I linked after all. Apologies. However, there are not great answers to that question, so any tips would be appreciated.
This line:
document.getElementById("test").innerHTML = this.responseText;
Is simply going to add the contents of this.responseText to the element test. I'm sure you understand that much, but what you want to do is completely wrong. The scripts within the output of get.php will NOT be executed at all and you have corrupted your HTML as well by including a second <html><head> and more.
A better approach would be if get.php returned the data from you DB query as a JSON string like so:
$data;
while ($row = mysqli_fetch_array($result)) {
$data[]['pic'] = $row['pathToPic'];
$data[]['wwwImg'] = substr($row['pathToPic'], 13);
}
echo json_encode($data);
exit;
And then back on gallery.php you do something like:
if (this.readyState == 4 && this.status == 200) {
formatResult(this.responseText);
}
// ... snip ...
function confirmJson(str) {
var j = null;
if (str) {
try { j = JSON.parse(str); }
catch (e) { j = null; }
}
return j;
}
function formatResult(str) {
var data = confirmJson(str);
if (! data) {
console.log("result is not JSON");
return;
}
// do stuff with the data
var i, max = data.length;
var h = document.createElement("div");
var img;
for(i=0;i<max;i++) {
img = document.createElement("src");
img.src = data[i].wwwImg;
img.addEventListener("click",function(){ alert("pic:["+ data[i].pic +"]"); },true);
h.appendChild(img);
}
document.getElementById("test").innerHTML = h;
}
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.
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.
I've a form where i have 4 buttons which are used for "insert, update, delete and retrieve" operations for a table. I can fill in the fields and click any button and respective operations take place. The DB operations takes place in PHP. But when the data is being displayed, it goes to a separate page. I want the table data to be displayed in the same page. I know it's possible using javascript or something but I'm pretty new to this coding. So im getting very confused. Have been trying for the past 3 days. Nothing worked out. If anyone could teach me clearly.
HTML:
<!DOCTYPE html>
<html>
<head>
<title>Untitled Document</title>
</head>
<body>
<h2>SELECT THE OPERATION YOU WANT TO PERFORM<h2>
<form method="post" action="open.php">
Id: <input type="text" name="Id" />
Name: <Input type="text" name="Name" />
BloodGroup: <input type="text" name="BloodGroup" /><br /><br />
<input type="submit" name="insert" value="Insert" />
<input type="submit" name="update" value="Update" />
<input type="submit" name="delete" value="Delete" />
<input type="submit" name="retrieve" value="retrieve" />
</form>
</body>
</html>
PHP:
<?php
$dbhost = 'localhost';
$dbuser = 'root';
$dbpass = '';
$dbname = 'DB';
$conn = new mysqli($dbhost, $dbuser, $dbpass, $dbname);
if(mysqli_connect_error())
{
die("couldn't connect" . $conn->connect_error());
}
$id = $_POST['Id'];
$name = $_POST['Name'];
$blood = $_POST['BloodGroup'];
if(isset($_POST['insert'])){
$insert = "Insert into ins(Id, name, BloodGroup) values ('$id','$name', '$blood')" ;
if($conn->query($insert) === TRUE) {
echo ("Input data entered successfully");
} else {
echo ("Input data failed to be entered" . $conn->error());
}
$conn->close();
} elseif(isset($_POST['update'])) {
$update = "update ins set Name='".$name."', BloodGroup='".$blood."' where Id='".$id."'";
mysql_query($update);
if($conn->query($update) === TRUE) {
echo ("Data updated successfully");
} else {
echo ("Data cant be updated" . $conn->error());
}
$conn->close();
} elseif(isset($_POST['delete'])) {
$id = $_POST['Id'];
$delete = "delete from ins where Id='".$id."'";
if($conn->query($delete) === TRUE) {
echo ("Data deleted successfully");
} else {
echo ("Data cant be updated" . $conn->error());
}
$conn->close();
}
else {
$id = $_POST['Id'];
$retrieve = "SELECT * FROM ins WHERE Id = ".$id;
if ($result=mysqli_query($conn,$retrieve))
{
while ($row=mysqli_fetch_row($result))
{
echo '<table>
<tr>
<td>Id</td>
<td>Name</td>
<td>Blood Group</td>
</tr>
<tr>
<td>'.$row[0].'</td>
<td>'.$row[1].'</td>
<td>'.$row[2].'</td>
</tr>
</table>';
}
mysqli_free_result($result);
}}
$conn->close();
?>
You can do it by using two method.
First: Create two file one for display your work another for operation.
when you submit your form it will go on operation page and when after opration is complete you have to set page location to working page like this.
header('location:working page.php');
Second: by using Ajax.
The easiest way to do this is that you should use ajax for this. On clicking each of the buttons send the action to separate files or methods. Retrieve whatever comes depending on the type of which action was submitted and successfully update the page.
For eg: If you are retrieving then on click of retrieve you send it to some page where you retrieve the details and return it to the page and just update it. It's pretty easy if you do some google on this. Make sure all your button actions go to proper respective page.
Hope this was helpful.
As you mentioned this is one example here: for retrieving(I expect you know how to call another file using ajax and return values.)
$.ajax({
url: 'retrieve.php',
type: 'post',
data: { id: id, name: name, blood_group: blood_group },
success: function(your_success_variable) {
Here your_success_variable can be an array or json whatever you wish to send from the other file back just don't forget to return
$('#blood_group').value(your_success_variable['blood_group]);
}
});
When the user clicks a SUBMIT-button, your FORM is submitted to open.php on your server. The browser is then taken to that page; displaying the output of your PHP script.
If you don't want to redirect the user, use AJAX in JavaScript as suggested above. There's a nice example at: How to make an AJAX call without jQuery?
The neatest way is to use jQuery, but it's more pedagogical to follow the plain JavaScript example. Essentially, you change your <input type="submit"> to <input type="button">, and add onClick="performAction('insert');" to each button (with the action string according to each operation). The function performAction(act) is declared in JavaScript in your HTML page, and creates an (asychronous!) AJAX call to your open.php script.
Right now you're submitting the FORM as POST-data. The AJAX call will have to pass the data as (URL-encoded!) GET-data, so performAction(act) appends the FORM data onto the call to open.php. (I presume it's possible to call AJAX with POST-data, and do all this in a more elegant way, but this will work. Hopefully a nice learning example.)
function performAction (act) {
var xmlhttp;
if (window.XMLHttpRequest) {
// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp = new XMLHttpRequest();
} else {
// code for IE6, IE5
xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange = function() {
if (xmlhttp.readyState == 4 ) {
if(xmlhttp.status == 200){
alert('Request OK, result was: ' + xmlhttp.responseText);
}
else if(xmlhttp.status == 400) {
alert('There was an error 400')
}
else {
alert('something else other than 200 was returned')
}
}
}
var GET_encoded_params = "action=" + act
+ "&Id=" + encodeURIComponent(document.forms[0].Id.value)
+ "&Name=" + encodeURIComponent(document.forms[0].Name.value)
+ "&BloodGroup=" + encodeURIComponent(document.forms[0].BloodGroup.value);
xmlhttp.open("GET", "open.php?" + GET_encoded_params, true);
xmlhttp.send();
}
Note that your PHP script might have to "get" the GET-data slightly different than you're getting the POST-data now, but that should be a minor change. You will also need to get the desired action (insert, update, delete, retrieve) from the new action parameter.
Code was borrowed from How to make an AJAX call without jQuery? and modified without actually testing. Beware of typos :) and have fun!
retrieve button can be placed in a new form where action will be on the same page or empty right now you have placed in a form where action is open.php so every action is performing on that particular page.This will definitely help you if any error occurs let me know.
else if(isset($_POST['retrieve'])) {
$id = $_POST['Id'];
$retrieve = "SELECT * FROM `ins` WHERE `Id` = '".$id."';
if ($result=mysqli_query($conn,$retrieve))
{
while ($row=mysqli_fetch_row($result))
{?>
<table>
<tr>
<td>Id</td>
<td>Name</td>
<td>Blood Group</td>
</tr>
<tr>
<td><?php echo $row[0];?></td>
<td><?php echo$row[1];?></td>
<td><?php echo$row[2];?></td>
</tr>
</table>
<?php }
mysqli_free_result($result);
}}?>
In your code you are directly printing the echo statement in the while loop.Just use a variable $msg which will store all the contents.So your code will be
<?php
$msg="";
$msg.= '<table>
<tr>
<td>Id</td>
<td>Name</td>
<td>Blood Group</td>
</tr>';
while ($row=mysqli_fetch_row($result))
{
$msg.=' <tr>
<td>'.$row[0].'</td>
<td>'.$row[1].'</td>
<td>'.$row[2].'</td>
</tr>;'
}
$msg.='</table>';
?>
After this in your html just echo the variable $msg
so your html code will be
<body>
<h1></h1>
<form>...</form>
<?php echo $msg;?>
</body>
I'm trying to teach myself how to use AJAX with PHP. However, I am unable to get my (likely broken) AJAX code to change the value of my $things variable. Instead, what happens is the index.php page gets reloaded and a brand new button gets added. Here is my code:
<?php
$things = 0;
?>
<!DOCTYPE html>
<html>
<head>
<title>testing</title>
<script>
function changeMe()
{
var xmlhttp;
if (window.XMLHttpRequest)
{
xmlhttp = new XMLHttpRequest();
}
xmlhttp.onreadystatechange=function()
{
if (xmlhttp.readyState==4 && xmlhttp.status==200)
{
document.getElementById("changetest").innerHTML=xmlhttp.responseText;
}
}
var things = "<?php echo $things; ?>";
things++;
xmlhttp.open("GET","index.php?things="+things,true);
xmlhttp.send();
}
</script>
</head>
<body>
<?php
echo "<p>Things = <span id=\"changetest\"" . $things . "</span></p>";
?>
<button type="button" onclick="changeMe()">Change Content</button>
</body>
</html>
Any clue as to where I'm messing up?
There are at least two problems I can see:
You are not using the variable you are sending back to the server $_GET['things'] anywhere. Instead you are always setting it to 0 so you will never see anything else;
You are posting to the original php file. That means that the response will be the complete html page, including the head, etc. You probably want a separate file to only send back what you really need.
An example of a separate php file would just be to echo out what was sent in:
<?php
echo 'I sent the number: ' . intval($_GET['things']);
And now you should make an ajax request to this file and not index.php.
You should remove the quotes around "<?php echo $things; ?>"; and make it var things = <?php echo "'".$things."'"; ?> instead.