I'm trying to ajaxlly upload a file including i want to add a post data with it
var xhr = this._xhrs[id] = new XMLHttpRequest();
var queryString = qq.obj2url(params, this._options.action);
xhr.open("POST", queryString, true);
xhr.setRequestHeader("X-Requested-With", "XMLHttpRequest");
xhr.setRequestHeader("X-File-Name", encodeURIComponent(name));
xhr.setRequestHeader("Content-Type", "application/octet-stream");
xhr.send(file);
how can i add x=y
as a post data ?
CORRECT NEW ANSWER
Look at the @habibutsu's answer below
WRONG OLD ANSWER
You can't post a file and a form data as x=y&w=z. They are two different Content Types.
For x=y you should use a content-type like this: application/x-www-form-urlencoded
I suggest you to split your AJAX request in two different or insert these data into the url: myUrl + 'query.php?x=y'.
Ciao
Wilk