Not able to call .php file through JQuery

Question

Below is my JQuery code. I just added an 'alert' to test whether data is retrieved correctly or not. But this code is not working. I think it is simply not able to get inside the file "filldistricts.php".

<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.3/jquery.js" />

<script type="text/javascript">
$(document).ready(function(){
    $("#inDistrict").change(function(){
    $.getJSON("filldistricts.php",function(j){
        alert(j[0]);
    })
  })
})
</script>

Below I am giving the code of "filldistricts.php":

<?php
    $query = "SELECT districtid FROM districtmaster";
    try
    {
        $result = mysql_query($query);
        $c = "";

        while($row = mysql_fetch_assoc($result))
        {
            $c = $c.$row['districtid'].",";
        }
        $c = $c."Select";
        echo json_encode(array($c));
    }
    catch(exception $ex)
    {
        echo "<script type='text/javascript'>alert('".$ex."');</script>";
    }
?>

Where is the problem?

Gersom · Accepted Answer · 2010-01-01 11:34:39Z

2

Maybe php is returning an exception? I thinks that doens't work with getJSON because it's not JSON..

answered Jan 1, 2010 at 11:34

Gersom

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1 Comment

RKh Over a year ago

There is a Try..Catch. Why I am not getting an exception alert?

mauris · Accepted Answer · 2010-01-01 11:35:15Z

0

Since you're using getJSON in jQuery, the server return type should be JSON string, thus you cannot put javascript and alert() when there's an error.

Thus instead of

echo "<script type='text/javascript'>alert('".$ex."');</script>";

use

echo json_encode(array($ex)) ;

that will allow you to see the exception properly.

answered Jan 1, 2010 at 11:35

mauris

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2 Comments

RKh Over a year ago

I removed the try..catch altogether and added error_reporting(E_ALL^E_NOTICE) on top.

mauris Over a year ago

did you try calling the php file from browser directly?

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