I am trying to make easy-to-use form for the users who are using our internal application. Basicly I have a select element. What I want is when I choose anything I want to fill specified text boxes.
Codes are below. That works fine but fill only 1 box.Pls do not forget normally I am getting data from the mySQL db.
html file:
<script src="choose.js"></script>
<select onchange="ChooseGuest(this.value)">
<option value="1">guest 1</option>
<option value="2">guest 2</option>
<option value="3">guest 3</option>
</select>
<input type="text" id="guestdetails"/>
<input type="text" id="guestdetails2"/>
php data file:
$AskForIt = $_GET["q"];
If ($AskForIt == "1") {
echo "guest 1 data";
} Elseif ($AskForIt == "2") {
echo "guest 2 data";
} Elseif ($AskForIt == "3") {
echo "guest 3 data";
}
js file:
var xmlHttp
function ChooseGuest(str)
{
xmlHttp=GetXmlHttpObject();
if (xmlHttp==null)
{
alert ("upgrade your browser");
return;
}
var url="guests.php";
url=url+"?q="+str;
url=url+"&sid="+Math.random();
xmlHttp.onreadystatechange=stateChanged;
xmlHttp.open("GET",url,true);
xmlHttp.send(null);
}
function stateChanged()
{
if (xmlHttp.readyState==4)
{
document.getElementById("guestdetails").value=xmlHttp.responseText;
}
}
function GetXmlHttpObject()
{
var xmlHttp=null;
try
{
// Firefox, Opera 8.0+, Safari
xmlHttp=new XMLHttpRequest();
}
catch (e)
{
// Internet Explorer
try
{
xmlHttp=new ActiveXObject("Msxml2.XMLHTTP");
}
catch (e)
{
xmlHttp=new ActiveXObject("Microsoft.XMLHTTP");
}
}
return xmlHttp;
}
From what I understand, you are trying to fill the input boxes with data pulled from your DB and have the appropriate data shown depending on which drop down is selected. PHP is server side only. Once the page loads the PHP will only run once and not again until the page is reloaded. So, you'll need to access those vars using JS. This is probably why only the first selection is being filled.
If this is so, you will need to go into your PHP file and create JS vars that are equal to the data accessed from the DB. Simply set your JS vars equal to a PHP echo of the vars from your php:
PHP Data File
<?php
var somePHPvar = "DB Value";
?>
<script>
var someVar = <?php echo somePHPvar; ?>;
</script>
If this is all done on the same PHP file, don't forget to include it on the page and call these JS vars. From there you'll be able to fill those values using JS.
Here is a way you can do this with just jQuery.
Example: https://jsfiddle.net/Twisty/u2gqwfL4/
JavaScript
var myData = [
"guest 1 data",
"guest 2 data",
"guest 3 data"
];
$(function() {
function ChooseGuest(v, str) {
// Example AJAX Post call in jQuery
// Replace '/echo/json/' with 'guests.php'
// or use $.get() in the same way
$.post("/echo/json/", {
json: JSON.stringify(myData),
q: str
}, function(results) {
console.log(results, v);
$("#guestdetails" + v).val(results[v - 1]);
});
}
$("#guestSelect option").click(function() {
var selVal = parseInt($(this).val());
var selText = $(this).html();
ChooseGuest(selVal, selText);
});
});
Your code will be a bit different due to the nature of jsfiddle and the way exampled are made. I would suggest:
function ChooseGuest(v, str) {
$.get("guests.php", {
q: str,
sid: Math.random()
}, function(results) {
console.log(results, v);
$("#guestdetails" + v).val(results);
});
}
Related
i am trying to create an ajax favorite button in php similar to instagram and twitter.
my code seems to be fine and i am doing everything correctly and tried to get it working this way:
php code:
$user_id = $_SESSION['active_user_id'];
extract($_POST);
extract($_GET);
if(isset($_GET['message']))
{
$id=$_GET['message'];
$q=$db->prepare("SELECT msgid,date,text
FROM messages
WHERE to_id=? and msgid=?");
$q->bindValue(1,$user_id);
$q->bindValue(2,$id);
$q->execute();
$row2=$q->fetch();
$d=$row2['date'];
$fav_questionq=$db->prepare("SELECT *
FROM messages
LEFT JOIN users
ON messages.to_id=users.id
WHERE users.id=? AND messages.msgid=?
");
$fav_questionq->bindValue(1,$user_id);
$fav_questionq->bindValue(2,$id);
$fav_questionq->execute();
$frow=$fav_questionq->fetch();
$fquestion= $frow['text'];
$result = $db->prepare("SELECT * FROM fav_messages
WHERE username=? AND message=?");
$result-bindValue(1,$user_id);
$result-bindValue(2,$id);
$result->execute();
if($result->rowCount()== 1 )
{
$deletequery=$db->prepare("DELETE FROM fav_messages WHERE message=?");
$deletequery->bindValue(1,$id);
$deletequery->execute();
echo("<script>location.href = 'index.php?a=recieved';</script>");
}
else
{
$insertquery = $db->prepare("INSERT INTO fav_messages (username,message,fav_question,fav_date) values(?,?,?,?)");
$insertquery->bindValue(1,$user_id);
$insertquery->bindValue(2,$id);
$insertquery->bindValue(3,$fquestion);
$insertquery->bindValue(4,$d);
$insertquery-execute();
}
echo("<script>location.href = 'index.php?a=recieved';</script>");
}
javascript code:
<script>
function GetXmlHttpObject() {
var xmlHttp=null;
try
{
// Firefox, Opera 8.0+, Safari
xmlHttp=new XMLHttpRequest();
}
catch (e)
{
// Internet Explorer
try
{ xmlHttp=new ActiveXObject("Msxml2.XMLHTTP"); }
catch (e)
{ xmlHttp=new ActiveXObject("Microsoft.XMLHTTP"); }
}
return xmlHttp;
}
function ajaxfav(){
var xmlHttp=GetXmlHttpObject();
var url="favorite.php?message="+document.msgidform.fav_message.value;
xmlHttp.onreadystatechange=function(){
if(xmlHttp.readyState==4){
alert("Message is favorited");
}
}
xmlHttp.open("GET",url,true);
xmlHttp.send(null);
}
</script>
HTML form and link code:
<form name="msgidform" method="post">
<input type="hidden" name="fav_message" id="" <?php echo "value= '$msg_id'"; ?>></p>
</form>
<a class="msg-icon" href="" onclick="ajaxfav();"><img
src="images/linedfav.png" id='img'></img></a>
but i kept getting an error in the console that reads:
"Uncaught TypeError: Cannot read property 'value' of undefined
at ajaxfav" but i've fixed it and now it immediately goes to the alert message, but nothing is inserted into the database, meaning that the data was not sent to the php file.
can someone advise me on what i can do?
network tab of the ajax call
please i would appreciate any help or suggestion. the php file is called but does not insert anything into the database.
You can use
document.msgidform.elements['fav_message'].value
in your ajaxfav() function to get name field value.
I am relatively new to PHP and I am (trying to) developing my first AJAX code...
I am trying to pass an attribute from a select with several options, each option with its very own unique attribute "values" that is a String.
The PHP file GETs the value and does something with it (for instance prints it out). And the generated file is then returned to my main HTML page.
Here is the source code...
Here is my javaScript:
function showUser(myElement) {
var str = myElement.options[myElement.selectedIndex].getAttribute("values");
if (str == "") {
document.getElementById("myResponse").innerHTML = "";
return;
} else {
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 && xmlhttp.status == 200) {
document.getElementById("myResponse").innerHTML = xmlhttp.responseText;
}
}
xmlhttp.open("GET","getuser.php?q="+encodeURIComponent(str),true);
xmlhttp.send();
}
}
My HTML is as following:
<form>
<select name="users" onchange="showUser(this)">
<option values="">Select a person:</option>
<option values="Rasmus Lerdorf">Creator of PHP</option>
<option values="Linus Torvalds">Developed Linux</option>
<option values="Dennis Ritchie">Developper of C</option>
</select>
</form>
<br>
<div id="myResponse"><b>Person info will be listed here...</b></div>
My PHP is:
<?php
$q = intval($_GET['q']);
echo $q;
?>
I get 0 instead of the text string, what is wrong with my code?
Silly me,
I copied and pasted a basic example from a tutorial and I did not realize that the PHP code was casting the string that was sent over the GET to an integer.
$q = intval($_GET['q']);
The code should be :
$q = $_GET['q'];
On my earlier stage of the development I was indeed passing an integer, so the code was working.
The code returned 0 instead of an error because the intval of a string is 0 on failure.
Writing down the question here I realized the mistake...
I have 2 php files, the first is BAConsult.php which is the main file, and the other is BAConsultRecordsAJAX.php. This line of code is in BAConsultRecordsAJAX.php:
while($row = mysqli_fetch_array($consultresult)) {
$skincareinuse=explode(",",$row['skincarecurrentlyinuse']);
}
In BAConsult.php I have this:
echo $skincareinuse;
Is it possible to bring over the $skincareinuse value from the BAConsultRecordsAJAX.php page, to the $skincareinuse variable on the BAConsult.php page?
Such that, lets say whenever I do a click on the table row, the value of $skincareinuse will be updated and shown by the echo, without the page refreshing?
[EDITED TO SHOW MORE OF MY CODES]
This is BAConsult.php file, where my main code is.
<?php echo $jsonvariable; ?>//Lets say i want to store the JSON data into this variable
function showconsultationdata(str) { //face e.g and checkboxes for that date selected.
if (str == "") {
document.getElementById("txtHint2").innerHTML = "";
return;
} else {
if (window.XMLHttpRequest) {
// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp = new XMLHttpRequest();
} else {
// code for IE6, IE5
xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange = function() {
if (xmlhttp.readyState == 4 && xmlhttp.status == 200) {
document.getElementById("txtHint2").innerHTML = xmlhttp.responseText;
var a = JSON.parse($(xmlhttp.responseText).filter('#arrayoutput').html());
$("textarea#skinconditionremarks").val(a.skinconditionremarks);
$("textarea#skincareremarks").val(a.skincareremarks);
var test = a.skincareinuse;
//I want to store this ^^ into the php variable of this page.
}
};
xmlhttp.open("GET","BAConsultRecordsAJAX.php?q="+str,true);
xmlhttp.send();
}
}
This is my BAConsultRecordsAJAX.php page.
$q = $_GET['q']; //get dateconsulted value
$consult="SELECT * FROM Counsel where nric='$_SESSION[nric]' and dateconsulted='$q'";
$consultresult = mysqli_query($dbconn,$consult);
while($row = mysqli_fetch_array($consultresult)) {
$skincareinuse=explode(",",$row['skincarecurrentlyinuse']);
$skincondition=explode(",",$row['skincondition']);
$queryResult[] = $row['skincareremarks'];
$queryResult[] = $row['skinconditionremarks'];
}
$skincareremarks = $queryResult[0];
$skinconditionremarks = $queryResult[1];
echo "<div id='arrayoutput'>";
echo json_encode(array('skincareremarks'=>$skincareremarks
'skinconditionremarks'=>$skinconditionremarks
'skincareinuse'=>$skincareinuse,
'skincondition'=>$skincondition));
echo "</div>";
You don't actually want to transfer the variable from BAConsultRecordsAJAX.php to BAConsult.php, but to BAConsult.js (I suppose that's the name of your JS file).
In reality, even that's what you wanted to do, it's impossible, because your main PHP is processed before the page even loads. What you can do, however, is overwrite the rendered value with a new one with the use of JavaScript.
To do that, send an AJAX request to BAConsultRecordsAJAX.php requesting the variable's value as shown below:
In JavaScript:
var xhttp = new XMLHttpRequest();
xhttp.onreadystatechange = function() {
// Check if the AJAX request was successful
if (xhttp.readyState === 4 && xhttp.status === 200) {
var td = document.getElementById("your table td value's id");
var td.innerHTML = xhttp.responseText; // gets the value echoed in the PHP file
}
xhttp.open("POST", "BAConsultRecordsAJAX.php", true);
xhttp.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
xhttp.send(""); // You don't need to send anything to the PHP file
In PHP:
<?php
echo $skincareinuse;
?>
[EDIT]:
If you are using inline JavaScript use the following code:
<script type = "application/javascript">
var xhttp = new XMLHttpRequest();
xhttp.onreadystatechange = function() {
// Check if the AJAX request was successful
if (xhttp.readyState === 4 && xhttp.status === 200) {
var td = document.getElementById("your table td value's id");
td.innerHTML = xhttp.responseText; // gets the value echoed in the PHP file
}
xhttp.open("POST", "BAConsultRecordsAJAX.php", true);
xhttp.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
xhttp.send(""); // You don't need to send anything to the PHP file
</script>
[EDIT 2]:
To pass a JSON object to PHP file via an AJAX request you have to make it a string. That requires the following code:
var JSONstring = JSON.stringify(yourJSONobject);
Then you can pass it in your PHP file with the AJAX request by putting it inside send() as shown:
xhttp.send("json_object=" + JSONstring);
Then in PHP, in order to use it, you have to decode it:
<?php
$json_object = json_decode($_POST["json_object"]);
// Now it's ready to use
?>
[About the code]:
Notes about the first file:
First of all, you have put your plain JavaScript code inside a PHP file and therefore it will not work. You have to wrap it in <script type = "application/javascript></script>
I don't have a clue what you are trying to do here:
Code:
var a = JSON.parse($(xmlhttp.responseText).filter('#arrayoutput').html());
It seems like you are trying to filter out and parse the innerHTML of #arrayoutput.
If the response is a JSON string, not HTML, so you absolutely can't do that. The logic of this above line is flawed.
How I would write your code:
function showconsultationdata(str) {
var xmlhttp;
if (!str) {
$("#txtHint2").empty();
} else {
xmlhttp = new XMLHttpRequest();
// Providing support for a 15-year-old browser in 2016 is unnecessary
xmlhttp.onreadystatechange = function() {
if (xmlhttp.readyState == 4 && xmlhttp.status == 200) {
$("#txtHint2").html(xmlhttp.responseText);
var a = JSON.parse(xmlhttp.responseText);
$("textarea#skinconditionremarks").val(a.skinconditionremarks);
$("textarea#skincareremarks").val(a.skincareremarks);
var test = a.skincareinuse; // The variable you want
// Its saved in "text" and can use something like:
// $("#element").html(test); to have it replace your current text
}
};
xmlhttp.open("GET","BAConsultRecordsAJAX.php?q="+str,true);
xmlhttp.send();
}
}
Notes about the second file:
I don't know if the query works for you, but it probably shouldn't, because:
$_SESSION is an associative array and you pass nric, whereas it should be "nric".
As $_SESSION is an array, the proper method to insert its value to your query is: {$_SESSION["nric"]}.
To ensure that you don't become the victim of an SQL Injection, use parameterised queries or at least sanitise somehow the received data, because GET is relatively easy to hack. Check the improved code later on how to do it.
How I would write your code:
$q = $_GET['q']; //get dateconsulted value
$dbconn = mysqli_connect("localhost", "root", "") or die("Error!");
mysqli_select_db($dbconn, "database") or die("Error!");
$consult = "SELECT * FROM Counsel where nric='{$_SESSION["nric"]}' and dateconsulted = ?";
if ($stmt = mysqli_prepare($dbconn, $consult)) { // By using a prepared statement
mysqli_stmt_bind_param($stmt, "s", $q); // you eliminate the chances to have
if (mysqli_stmt_execute($stmt)) { // an SQL Injection break your database
$consultresult = mysqli_stmt_get_result($stmt);
if (mysqli_num_rows($consultresult) > 0) {
while ($row = mysqli_fetch_array($consultresult)) {
$skincareinuse = explode(",",$row['skincarecurrentlyinuse']);
$skincondition = explode(",",$row['skincondition']);
$queryResult[] = $row['skincareremarks'];
$queryResult[] = $row['skinconditionremarks'];
}
$skincareremarks = $queryResult[0];
$skinconditionremarks = $queryResult[1];
$array = array('skincareremarks'=>$skincareremarks
'skinconditionremarks'=>$skinconditionremarks
'skincareinuse'=>$skincareinuse,
'skincondition'=>$skincondition);
echo "<div id='arrayoutput'>";
echo json_encode($array);
echo "</div>";
mysqli_stmt_close($stmt);
mysqli_close($dbconn);
exit;
}
}
}
Obviously you may change the values to suit your requirements, you must have jquery installed, the time value could be anything you want
$.ajaxSetup({ cache: false });
setInterval(function() {
var url = "pagethatrequiresrefreshing.php";
$('#divtorefresh').load(url);
}, 4000);
Using include will most likely work. From the docs:
When a file is included, the code it contains inherits the variable scope of the line on which the include occurs. Any variables available at that line in the calling file will be available within the called file, from that point forward.
BAConsultRecordsAJAX.php
while($row = mysqli_fetch_array($consultresult)) {
$skincareinuse[] = explode(",",$row['skincarecurrentlyinuse']);
}
include "BAConsult.php";
Below is my textbox code
<input id="society_name" onBlur="showsociety(this.value)" />
<input id="societyid" name="society" />
Below is my javascript which call addressdata.php page...
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.10.1/jquery.min.js" type="text/javascript"></script>
<script>
function showsociety(str)
{
if (window.XMLHttpRequest)
{ xmlhttp=new XMLHttpRequest();}
else
{ xmlhttp=new ActiveXObject("Microsoft.XMLHTTP"); }
xmlhttp.onreadystatechange=function()
{
if (xmlhttp.readyState==4 && xmlhttp.status==200)
{
var data = JSON.parse(xmlhttp.responseText);
for(var i=0;i<data.length;i++)
{
document.getElementById("societyid").value = data[i].societyid;
}
}
}
xmlhttp.open("GET","addressdata.php?q="+str,true);
xmlhttp.send();
}
</script>
Addressdata.php page
<?php
require_once('includes/config.php');
$q = $_GET['q'];
$city = $database->getRows("SELECT SM.id AS societyid,SM.society from societymaster SM WHERE SM.society = :society", array(':society'=>"$q"));
$info = array();
foreach($city as $row)
{
$cID = $row['societyid'];
$info[] = array('societyid' => $cID);
}
echo json_encode($info);
?>
I need to fetch id in multiple textbox like above given ex...in my form.
So is this possible to convert all php code to function in addressdata.php and call this function only from javascript...
FOR EX - i need to make whole php code of addressdata.php file as it is in function and call tis with below javascript on textbox blur event..
If I understood you correctly you want to add more text input elements into your page and be able to use this whole process of showing society on each of this elements.
The problem is not converting php code into a function (which would bring nothing).
What you want is to be able to tell showsociety() function which input element should it work on.
In the easiest case you can add additional parameter to the fucntion:
showsociety(str, id) {...}
And use this ID to search for correct element on the page.
[...]
document.getElementById(id).value = data[i].societyid;
[...]
.
<input id="society_name" onBlur="showsociety(this.value, this.id)" />
It can be done better but I think with such simple solution you should not have much problems.
Hope it helped.
I have that
JavaScript
flagQuery = session.reponseText;
alert(flagQuery);
flagQuery = JSON.parse(flagQuery);
contentElt.innerHTML = "valeur FLAG" + flagQuery;
My javascript must receive a flag sent from php.
The code above is the treatment of flag when it is received.
In my php I declare the variable flag to true.
I want to treat this with Ajax, if it's possible.
I don't know what to add in the PHP to get the flag in JavaScript?
I should prefer not to have to overload my code with a jQuery bookshop that I will use it for half a score line of code.
I assume want to access php value through ajax request. And you can do that liek below;
PHP: flag.php
<?php
$flag = "Your flag value";
echo $flag;
JS:
$.ajax({
url: "flag.php",
type: "get",
success: function(data){
alert("Flag value is: " + data);
},
error:function(){
alert("Error occured!");
}
});
Pure JS:
<script>
function getFlagValue() {
var xmlHttp;
if (window.XMLHttpRequest) {
xmlHttp = new XMLHttpRequest();
} else {
xmlHttp = new ActiveXObject("Microsoft.XMLHTTP");
}
xmlHttp.onreadystatechange = function() {
if (xmlHttp.readyState==4 && xmlHttp.status==200) {
alert(xmlHttp.responseText);
}
}
xmlHttp.open("GET","flag.php",true);
xmlHttp.send();
}
</script>
Hello. Try this and see if it's you need.
//PHP File
<?php
echo json_encode(array("one","two"));
?>
//Js File
$.get("page.php", function(data){
alert(data)
})