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jQuery intercept form submission to param string

I have a form. This form submits via POST to an iframe, that, in turn, has the request processed and, depending on the result, a javascript is executed that changes the parent's content according the the submission's validity.

Now, I don't like this procedure. I want the ability to submit several forms simultaneously, but I have only this one hidden iframe. So I would like to do it with AJAX, creating a separate request for each form submission.

However, my form is complicated; it consists of checkboxes that add variables to arrays, of images that are clicked and whose click coordinates I need sent, and complicated stuff like that - which is why I, instead of calculating each the value of each input and adding it to a post parameter string (by the way: I don't know how I can create arrays that way), I would much prefer to rather have the submission content intercepted, saved as a post string with all those parameters, and then use this string for the AJAX POST request.

That I would like to do in this function:

$('#myForm').submit(function(event){

    // process the submission, e. g. event.getContent().toPostString();

    // create the AJAX request and send it and attach listeners (I know how to do this line ;)

    return false; // I don't want the form submitted (to the iframe)

});

Thanks in advance!

like image 481
arik Avatar asked Jun 15 '11 19:06

arik


2 Answers

Don't use an iframe, just use jQuery's ajax methods: (I used post() in the example below)

$('#myForm').submit(function(event){

             //url          //post data
    $.post(this.action, $(this).serialize(), function(returnData){
           //do something with returnData
    })

    return false; //do not submit form the normal way

});

Here is an example: http://jsfiddle.net/maniator/Y6r8E/
Type something into the form and submit it.

like image 104
Naftali Avatar answered Oct 17 '22 03:10

Naftali


jQuery's serialize() function will gather the form data for you, to allow for easy form submission via .ajax() or .post().

like image 35
George Cummins Avatar answered Oct 17 '22 03:10

George Cummins