jQuery AJAX call won't work

3 Posts

StyloMilo Reply #1, 15 years, 8 months ago

Hi there,

For your information... I’m using MODx 2.0.7 traditional and jQuery.

I ran into the following problem: I have a form on my page and when the page unloads, I want to save the values of the form fields in the session. On other websites using another CMS I did this: On unload I call jQuery’s ajax-function to send the values of the form fields to a PHP-file using the POST-method. the PHP-file saves the values in the $_SESSION-array and echos the success or failure back to the jQuery part. Everything works fine there. But now on my MODx-site i can’t get this to work right. I don’t get any error message nor is the success part of the ajax-call triggered. Just nothing happens. When I replace the $.ajax(...) with alert(’something’), I get the alert so the error has to be in the $.ajax-part. I hope I made my problem clear so everyone can understand it. Here are the bits of code I use...

The jQuery-AJAX-call:

$(window).unload(function() {   
    $.ajax({
        type: "POST",
        url: "http://www.example.com/modx/saveToSession.php",
        data: {
	    formfieldvalue1: $('[name=formfield1]').val(), 
	},
	success: function(msg){
            alert('bla');
	}
    });
});

The called PHP file:

session_start();
if($_POST) {
    $_SESSION['formvalues'] = '';	
    $formvalues = array($_POST['formfieldvalue1']);
    $_SESSION['formvalues'] = $formvalues;
    $data['success'] = true;
} else {
    $data['success'] = false;
}
echo json_encode($data);

The strange thing is, that even when I replace the content of the PHP-file with something like "echo ’success’;", the success-part in the AJAX-call is still not triggered. I’m very confused and I hope somebody out there is able to help me solve this, cause this has been driving me crazy for days now.

Thanks a lot...
Joachim

Windschutznetz | PVC-Streifenvorhang | Druckverschlussbeutel

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5,888 Posts

Bruno17 Reply #2, 15 years, 8 months ago

I would try to put your php-code into a snippet without session_start().

then put a snippet-call of that snippet to a new resource with blank template.

call the url of that resource in your ajax-request.

-------------------------------

you can buy me a beer, if you like MIGX

http://webcmsolutions.de/migx.html

Thanks!

1,336 Posts

shamblett Reply #3, 15 years, 8 months ago

Put an error function in your AJAX handler as well as the success one, see if this is triggered, jQuery API refers e.g

error(jqXHR, textStatus, errorThrown)Function
A function to be called if the request fails......

Use MODx, or the cat gets it!

3 Posts

StyloMilo Reply #4, 15 years, 8 months ago

Hi,

I tried what y’all suggested, but I’m not any closer to getting this to work.

I removed every JavaScript from my page and changed the AJAX-part to the following:

$(document).ready(function() {
    $(window).unload(function() {   
        $.ajax({
            url: "[[++site_url]][[~26]]",
            statusCode: {
                404: function() {
                    alert('page not found');
                },
                200: function() {
                    alert('page found');
                }
            },
            type: "POST",
            data: {
	        formfielvalue1: $('[name=formfield1]').val() 
	    },
            success: function(){
	        alert('success');
	    },
            error: function(){
                alert('error');
            }
        });
    });
});

I put PHP-code into a snippet and made a document (ID:26) calling this snippet like that:

[[!saveToSession]]

In order to minimize potential sources of errors I just have the following in the snippet:

echo 'thisIsTheReturnvalueOfTheSnippet';

But anyways... I don’t get any status alert, success or error message. It just doesn’t do anything. Does anybody know what I am missing? I just don’t get the mistake, as this works on other websites with other CMS.

Windschutznetz | PVC-Streifenvorhang | Druckverschlussbeutel

www.verpackung-tvv.de

5,888 Posts

Bruno17 Reply #5, 15 years, 8 months ago

try this one and you will see, when you are waiting long enough, there comes an answer. (testet!)

$(document).ready(function() {
    $(window).unload(function() {   
        $.ajax({
            url: "[[++site_url]][[~26]]",
            statusCode: {
                404: function() {
                    alert('page not found');
                },
                200: function() {
                    alert('page found');
                }
            },
            type: "POST",
            data: {
	        formfielvalue1: $('[name=formfield1]').val() 
	    },
            success: function(){
	        alert('success');
	    },
            error: function(){
                alert('error');
            }
        });
        alert('Ich warte erstmal noch auf Antwort von [[++site_url]][[~26]]');
    });
});

-------------------------------

you can buy me a beer, if you like MIGX

http://webcmsolutions.de/migx.html

Thanks!

3 Posts

StyloMilo Reply #6, 15 years, 8 months ago

D’oh! That works... which means that the response takes too long and the browser leaves the page before the ajax-call is completed.

And for those of you experiencing the same problem: I solved now by adding the parameter

async: false

to the $.ajax-function. Like that the page is not left until the ajax-call is finished and the success-branch is executed.

Thanks a lot Bruno, for showing me the right direction!

Windschutznetz | PVC-Streifenvorhang | Druckverschlussbeutel

www.verpackung-tvv.de