jQuery $.get() not running php script

Question

I'm unable to run my php script at the click of the button.

If I run the script via the terminal, 'php get_funny_status.php' it returns the correct output. However, I'm not able to do this via AJAX. The beginning ajax alert shows up, but I'm not getting any responseText.

Am I missing something?

EDIT: I am testing this application view Preview Browser in Adobe Dreamweaver (I'm not sure if this has anything to do with the issue)

<script>

$( document ).ready(function() {

    $( ".nextStatus" ).click(function() {

        alert('beginning ajax');

        var statusNumber = '5';

        $.get('get_funny_status.php?' + statusNumber, function(responseText) {
            alert(responseText);
        });

    });
});

</script>

Here's my php script:

  <?php

    //initialize DB stuff

    $status_text_query = "SELECT * FROM funny_status WHERE STATUS_NUM = '". $_GET['statusNumber']."'";

    $result = mysql_query($status_text_query);

    while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {

        echo 'num likes: ' . $row['NUM_LIKES'];
        echo 'num dislikes: ' . $row['NUM_DISLIKES'];
        echo 'status text: ' . $row['STATUS_TEXT'];
        echo 'status num: ' . $row['STATUS_NUM'];   

     }  
 ?>

If I were you, I would use mysqli or PDO instead of mysql_ functions. The first line is a hacker delight, prone to SQL Injection. — Robson França
– Robson França, Commented Feb 8, 2014 at 1:07
It looks like yo have some good answers to work from below. I should also point out you have a HUGE SQL injection vulnerability that you should think about fixing. — Mike Brant
– Mike Brant, Commented Feb 8, 2014 at 1:08
NONE of these answers are helping. I already realized the syntax errors and TRIPLE CHECKED the path name. Could it have something to do with running a .php from Adobe Dreamweaver? I'm able to run the script from my terminal just fine. Just not through Dreamweaver! — Mark Kennedy
– Mark Kennedy, Commented Feb 8, 2014 at 1:43
Can you use XAMPP or something similar to that to see if it works? Also, you didn't mention an error log; does it get populated with anything when you run the script? — Chris Forrence
– Chris Forrence, Commented Feb 8, 2014 at 13:45

DeepBlue · Accepted Answer · 2014-02-08 01:03:48Z

1

Fix this from:

    $.get('get_funny_status.php?statusNumber', function(responseText) {
        alert(responseText);
    });

to

    $.get('get_funny_status.php?statusNumber='+statusNumber, function(responseText) {
        alert(responseText);
    });

answered Feb 8, 2014 at 1:03

DeepBlue

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Comments

Community · Accepted Answer · 2017-05-23 12:20:40Z

0

There are a few issues that I see:

In your PHP, _GET should be $_GET.

$status_text_query = "SELECT * FROM funny_status WHERE STATUS_NUM = '"
    . $_GET['statusNumber']."'";

In your AJAX call, you need to send your parameter ?statusNumber=' + statusNumber, ...

$.get('get_funny_status.php?statusNumber=' + statusNumber, 
    function(responseText) {

Not necessarily a syntax error: I noticed that you're using mysql_* to access the database. This is a bad idea; I'd advise switching to either the mysqli_* functions or to the PDO class. Related to this issue, if I were to set statusNumber to something like "'; DELETE FROM funny_status;--", I would be able to delete all of the data in that table. This is called SQL Injection, and it is a serious security issue.

edited May 23, 2017 at 12:20

CommunityBot

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answered Feb 8, 2014 at 1:04

Chris Forrence

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