JQuery: 'Uncaught TypeError: Illegal invocation' at ajax request - several elements

Question

I have two select elements, A and B: when A\'s selected option changes, B\'s options must be updated accordingly. Each element in A implies many elements in B, it\'s a one-t

Answer 1

From the jQuery docs for processData:

processData Boolean
^{Default: true}
By default, data passed in to the data option as an object (technically, anything other than a string) will be processed and transformed into a query string, fitting to the default content-type "application/x-www-form-urlencoded". If you want to send a DOMDocument, or other non-processed data, set this option to false.

Source: http://api.jquery.com/jquery.ajax

Looks like you are going to have to use processData to send your data to the server, or modify your php script to support querystring encoded parameters.

Answer 2

I've read in JQuery docs that data can be an array (key value pairs). I get the error if I put:

This is object not an array:

var data = {
        'mode': 'filter_city',
        'id_A': e[e.selectedIndex]
};

You probably want:

var data = [{
        'mode': 'filter_city',
        'id_A': e[e.selectedIndex]
}];

Answer 3

Try This:

            $.ajax({
                    url:"",
                    type: "POST",
                    data: new FormData($('#uploadDatabaseForm')[0]),
                    contentType:false,
                    cache: false,
                    processData:false,
                    success:function (msg) {}
                  });

Answer 4

function do_ajax(elem, mydata, filename)
{
    $.ajax({
        url: filename,
        context: elem,
        data: mydata,
        **contentType: false,
        processData: false**
        datatype: "html",
        success: function (data, textStatus, xhr) {
            elem.innerHTML = data;
        }
    });
}

Answer 5

I was getting this error while posting a FormData object because I was not setting up the ajax call correctly. Setup below fixed my issue.

var myformData = new FormData();        
myformData.append('leadid', $("#leadid").val());
myformData.append('date', $(this).val());
myformData.append('time', $(e.target).prev().val());

$.ajax({
    method: 'post',
    processData: false,
    contentType: false,
    cache: false,
    data: myformData,
    enctype: 'multipart/form-data',
    url: 'include/ajax.php',
    success: function (response) {
        $("#subform").html(response).delay(4000).hide(1); 
    }
});

Answer 6

$.ajax({
                    url:"",
                    type: "POST",
                    data: new FormData($('#uploadDatabaseForm')[0]),
                    contentType:false,
                    cache: false,
                    processData:false,
                    success:function (msg) {}
                  });