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jQuery AJAX and PHP variables

in Using jQuery • 1 year ago

Hey guys, I have a quick question.

I am using jQuery AJAX to login and logout. When the user is logged in, I want to show the user's username, which I store in a session. So, $_SESSION['username'] is equal to a given username. I am pretty new to jQuery, so so far, I have only displayed information that I have echoed out in a PHP script. Is it possible to somehow pass other information back to jQuery from the PHP script. Say I have a login.php and somepage.php. Somepage.php has the login form and uses login.php to handle the login process. Then the form on somepage.php is updated depending on the login result. Is there any way to pass the username back to somepage.php without making a script that simply echoes out the username?

Thanks in advance! :)

Andy

Replies(9)

kboudloche

Re: jQuery AJAX and PHP variables

1 year ago

you can pass pretty much anything as the result of an ajax call. just echo it in the php script that the ajax call is hitting, then parse it in the ajax success function.

-- Kevin

------------------------------------------------
http://www.tentonaxe.com - New Design!

andytk

Re: jQuery AJAX and PHP variables

1 year ago

Can you please provide a simple example of how I can accomplish what I wrote in the first post (or something where the principle is the same or similar)? Thank you.

wavetydes..

Re: jQuery AJAX and PHP variables

1 year ago

If the username is all your after the super simple way is make a getUsername.php file that looks like this:

<?php
session_start();
print $_SESSION['username'];
?>

Then, you could simply load that into where you want it your somepage (i.e):

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">
<html>
<head>
<title>somepage</title>
<script type="text/javascript" src="jquery.js"></script>
<script>
$(document).ready(function() {
$('#username').load('getUsername.php');
});
</script>
</head>
<body>
<div id="username"></div>
</body>
</html>

You could then access that value in your jQuery like so:

$('#username').html();

Of course this is just a sample and deals nothing with security...

andytk

Re: jQuery AJAX and PHP variables

1 year ago

Yes, I was thinking that it could be done like that. The reason why I made this topic was to find out if there are other ways to do it. This way is very straight forward, but I imagine that one can quickly have a whole lot of those small PHP scripts if multiple values are needed. I will do something similar to what you posted for the username, but is there another way to do it (particularly for returning multiple values)? I really just want to get good habits from the start. :-)

Thank you for your reply, I appreciate the help.

wavetydes..

Re: jQuery AJAX and PHP variables

1 year ago

For multiple values, working with JSON data objects can be pretty straight forward. Have your login.php return JSON values:

<?php
$data = array("username" => $_SESSION['username'],"nextvar" =>"value");
print json_encode($data);
?>

and process those back in your somepage via parseJSON. I'm not sure how you are posting your form, but this way would require that you do it via jQuery, and then do something with the data in a callback.

I.E.

var userInfo = jQuery.parseJSON(response);
alert(userInfo.username);

andytk

Re: jQuery AJAX and PHP variables

1 year ago

Great, that looks pretty simple. Thank you very much for your help, Tyler.

andytk

Re: jQuery AJAX and PHP variables

1 year ago

Hello again. As (somewhat) expected, I couldn't get this to work. Maybe I misunderstood something, but I did the following:

$.ajax({
url: 'php/login.php',
type: 'POST',
dataType: 'json',
data: 'user=' + $('#user').val() + '&pass=' + $('#pass').val(),
success: function(result) {
$('#load').fadeOut(250, function() { // Fading out loading gif
$(this).remove(); // Removing it
if (result.success) {
console.log("we succeeded, man");
}
else {
console.log("login failed");
}
});
}
});

And the following is my login.php:

if (login was successful) {
$data = array("username" => $_SESSION['username'], "success" => true);
echo json_encode($data);
}
else {
$data = array("username" => null, "success" => false);
echo json_encode($data);
}

I was using a friend's PHP server but it is using an old version of PHP, so the json_encode() method is not yet supported. I then switched to a server running PHP 5.2.13, but that gave me another problem; it just appears to just time out when posting to the PHP script. I am not getting a response in Firebug and after ~21 seconds, it just starts to count errors like crazy for an eternity:

I guess I should check whether the result is null or not, but that still doesn't explain why it seems to time out. Is there anything on the web server's part that could do this, such as a missing feature, low version of something or similar?

Thanks.

wavetydes..

Re: jQuery AJAX and PHP variables

1 year ago

Andy,

I ran your code (from the post) with the necessary adjustments (filling in missing code) on my system, and it functions normally in all of my browsers with 0 errors in Firefox. So, the potential problems are:

Accessing 'php/login.php'... perhaps its not even reaching this file and then simply timing out.
PHP Version: You said it was 5.2.13? Run a phpinfo() and search for json just to make sure.
Your JS is expecting JSON as its ONLY returning values, make sure nothing else in your login.php is outputting anything prior (or after) your echo of the JSON and screwing it up.

Here are my test files in case you can use them to test:

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">
<html>
<head>
<title>Untitled</title>
</head>
<body>
<script type="text/javascript" language="JavaScript1.2" src="jquery.js"></script>
<script>
$(document).ready(function() {
$.ajax({
url: 'test.php',
type: 'POST',
dataType: 'json',
data: 'user=' + $('#user').val() + '&pass=' + $('#pass').val(),
success: function(result) {
$('#load').fadeOut(250, function() { // Fading out loading gif
$(this).remove(); // Removing it
if (result.success) {
console.log("we succeeded, man: " + result.username);
$('#output').html("we succeeded, man: " + result.username)
}
else {
console.log("login failed: " + result.username);
$('#output').html("login failed: " + result.username)
}
});
}
});
});
</script>
<input type="text" name="user" id="user" value="Test User">
<input type="password" name="pass" id="pass" value="somepass">
<div id="load"><img src="spinner.gif"></div>
<div id="output"></div>
</body>
</html>

and the php

<?php
$login = true;
if ($login) {
$data = array("username" => "Test User", "success" => true);
echo json_encode($data);
}else{
$data = array("username" => null, "success" => false);
echo json_encode($data);
}
?>

Do {offer.assistance} while (questions > 1)

kboudloche

Re: jQuery AJAX and PHP variables

1 year ago

you don't even have to use jQuery.parseJSON if you use .ajax() and a dataType of "json" or .getJSON(), however their functionality is different from .load(), you would have to append the result to the target div within the success function.

http://api.jquery.com/jQuery.ajax/ http://api.jquery.com/jQuery.getJSON/

-- Kevin

------------------------------------------------
http://www.tentonaxe.com - New Design!

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