jQuery Ajax Form Data

Question

I have an Ajax form. i am caputuring the ID of the form. and when i the ajax call is a success. i want to pass this Id form BeforeSubmit to success. so that i can append the results to a place where i want.

here is the code

function StatusComments() {

    $('.status-comment').submit(function() {
        $(this).ajaxSubmit(options);
        return false;
    });

    var options = {
        beforeSubmit: showRequest,
        success: showResponse,
        resetForm: true
    };

    function showRequest(formData, jqForm, options) {
  var formID = $(this).attr("id");

    }

    function showResponse(responseText, statusText, xhr, $form) {
  alert(responseText);
  var formID = $(this).attr("id");
  alert(formID);
    }

}

Darin Dimitrov · Accepted Answer · 2010-11-17 06:52:11Z

4

The form is passed as argument to the success handler:

var formID = $form.attr('id');

Also I notice that you are using the jquery form plugin and still subscribing to the .submit event of the form which is not necessary:

$(function() {
    var options = {
        beforeSubmit: showRequest,
        success: showResponse,
        resetForm: true
    };
    $('.status-comment').ajaxForm(options);
});

function showRequest(formData, jqForm, options) {

}

function showResponse(responseText, statusText, xhr, form) {
    var formID = form.attr('id');
    alert(formID);
}

edited Nov 17, 2010 at 6:52

answered Nov 17, 2010 at 6:46

Darin Dimitrov

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16 Comments

Harsha M V Over a year ago

@Darin: i tried with the AjaxSubmit. didnt get it to work :( i have many .status-comment on the page. i want to pick up the one i clicked

Harsha M V Over a year ago

thanks. when ever i use the ajaxForm alone. i am not able to trigger the success function. so i am using with submit. i dont get any error too

Darin Dimitrov Over a year ago

Well, there must something wrong with your code because that's the way it is supposed to be used. It's much easier.

Harsha M V Over a year ago

ok. is it bcos. i am redenering the html at the form execution end ?

Darin Dimitrov Over a year ago

Yes, it's exactly because of this. When you call .ajaxForm, the form must be present in the DOM. So if it is a dynamically generated form you need to call the .ajaxForm after it is inserted in the DOM.

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