Jquery Ajax Data Post Problem in PHP

Question

I am using jquery's AJAX in my project. Today, I used it somewhere else with all same themethods but it doesn't work.

Is there something wrong with my script?

HTML:

<a class='btn edit_receipe_btn' id='myreceipe-52'>Edit</a>

JQuery:

(Click function works. When I put alert(instance) after var instance line, it works)

$(document).ready(function(){
$('.edit_receipe_btn').click(function(){
   var instance = $(this).attr('id');
   var dataString = 'process=userReceipeEdit&instance='+instance;
   $.ajax({
    type: 'POST',
    url: 'ajax/ajaxs.php',
    data: dataString,
    cache: false,
    success: function(msg) {
        alert(msg);
    }
    });
});
});

PHP:

$prcs = $_POST['process'];
if($prcs=='userReceipeEdit'){
        $instance = $_POST['instance'];
        return $instance;
    }

It appears the problem is in the PHP. What am I doing wrong?

You do have $prcs = $_POST['process']; in your PHP as well, yes? — sje397
– sje397, Commented Dec 13, 2010 at 15:25
Yeah, I just added it to my post. It was there from the beginning in my code. — aiternal
– aiternal, Commented Dec 13, 2010 at 15:26
if you take out that if $prcs statement can you hit the page directly in a browser? — wajiw
– wajiw, Commented Dec 13, 2010 at 15:28
Please don't put 'solved' in the title. If something someone else wrote solved your problem, please upvote and accept their answer. — George Stocker
– George Stocker, Commented Dec 13, 2010 at 16:04

Jasper De Bruijn · Accepted Answer · 2010-12-13 16:33:26Z

3

Is that the entire PHP page? if so, you should echo instead of return

answered Dec 13, 2010 at 16:33

Jasper De Bruijn

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Comments

aiternal · Accepted Answer · 2010-12-13 16:09:52Z

2

As Jasper De Bruijn notified me, the problem was in my php script. I should use echo instead of return: Wrong usage:

$prcs = $_POST['process'];
if($prcs=='userReceipeEdit'){
        $instance = $_POST['instance'];
        return $instance;
    }

Correct usage:

$prcs = $_POST['process'];
if($prcs=='userReceipeEdit'){
        $instance = $_POST['instance'];
        echo $instance;
    }

edited Dec 13, 2010 at 16:09

answered Dec 13, 2010 at 15:53

aiternal

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Comments

Jason · Accepted Answer · 2010-12-13 15:32:00Z

0

Two things, output the error and check to make sure instance is not undefined.

Output the error like this:

$('.edit_receipe_btn').click(function(){
   var instance = $(this).attr('id');
   var dataString = 'process=userReceipeEdit&instance='+instance;
   $.ajax({
    type: 'POST',
    url: 'ajax/ajaxs.php',
    data: dataString,
    cache: false,
    success: function(msg) {
        alert(msg);
    },
    error: function(response, status, error)
    { 
        alert(response.responseText);
        alert(status);
        alert(error);
    }
    });
});

answered Dec 13, 2010 at 15:32

Jason

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Comments

CrazyDart · Accepted Answer · 2010-12-13 15:32:05Z

0

I think you have formed this POST like a get, not sure why. Try doing this in JS:

var dataString = 
    {
        "process" : "userReceipeEdit",
        "instance" : instance
    };

answered Dec 13, 2010 at 15:32

CrazyDart

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