Solved

jquery .ajax() not executing url paramater

Posted on 2012-04-05
6
449 Views
Last Modified: 2012-04-18
Hi,

I have this bit of code to check if a username and/or email already exists in the database:
/**************************
* Check if usernamer or email exists in db already.
**************************/
function checkUsernameEmail(username, email) {
    $.ajax({
        type : 'POST',
        url: './model/modCheckUsernameEmail.php',
        dataType: 'json',
        data: {
            username : username,
            email : email
        },
        success: function(retData, textStatus, jqXHR) {
            if (retData.success==1) { //means does not username and/or email exits in database already
                return true;
            } else { //data.success==0 (i.e. username and/or email already exists in database)
                if (retData.username != '' || retData.username != null) {$('#usernameerror').text('username already exists in database').show(500);}  
                if (retData.email != '' || retData.email != null) {$('#emailerror').text('email already exists in database').show(500);}
                return false;
            }
        },
        error: function(XMLHttpRequest, textStatus, errorThrown) {
            alert(textStatus);
        }
    });
}

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It works fine on the first run. For instance, if I have a username and/or email in the database then it returns the error. But if I change the username and then click on the submit button again. The url parameter in the .ajax() is not executed. This line:
url: './model/modCheckUsernameEmail.php',

I can tell because my PHP debugger does not get stepped through to this file. Oddly, when I'm stepping through with firebug on the second submit with a different username I can see the variable is passed with this new username in this section:
data: {
            username : username,
            email : email
        },

But the url parameter is not called. Why is that and how can I work around this issue?

Thanks,
Victor
0
Comment
Question by:Victor Kimura
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6 Comments
 
LVL 20

Assisted Solution

by:Proculopsis
Proculopsis earned 500 total points
ID: 37812550

Could it be a cache problem, try adding the follwing:

cache: false
0
 
LVL 14

Expert Comment

by:binaryevo
ID: 37812574
I was thinking the same thing as Proculopsis...  Do as suggested and see what happens.
0
 

Author Comment

by:Victor Kimura
ID: 37812610
It doesn't work. =(

Any think else?
0
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LVL 20

Assisted Solution

by:Proculopsis
Proculopsis earned 500 total points
ID: 37813318
Just to make sure, add the following to the data:

random: new Date().getTime()
0
 

Accepted Solution

by:
Victor Kimura earned 0 total points
ID: 37845335
Hi fellas, I think what was happening is that I was calling checkUsernameEmail() from another .ajax({.success: function() {...}}); call and the initial calling function was not returning any value. So on the calling function, say, callingFunction() {.ajax({...}) I would call checkUsernameEmail() on success from within callingFunction() and this works. I think what was tripping me was that .ajax does not return any thing like:
var retVar = callingFunction() {.ajax({...})
doesn't work because retVar is null or ''.

Thank you for your input and help.
0
 

Author Closing Comment

by:Victor Kimura
ID: 37859798
Please read my last comment. =)
0

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