AJAX Submit the My Account Page
nicholas.alipaz - March 28, 2009 - 05:21
I am trying to submit this page via an ajax submit, but it never seems to post the values. I have attempted the following code which seems to operate correctly, however my account never changes. Any assistance would be great!
// initialize vars
var form_url = "http://mysite.com/user/me/edit";
var poststr="";
var http_request=false
var self=this;
// AJAX posting function
function makePOSTRequest(url, parameters, mode) { //mode 1=fetch page first
http_request = false;
if (window.XMLHttpRequest) { // Mozilla, Safari,...
http_request = new XMLHttpRequest();
if (http_request.overrideMimeType) {
switch(mode) {
case 1: http_request.overrideMimeType('text/html');
case 2: http_request.overrideMimeType('text/xml');
default: break;
}
// set type accordingly to anticipated content type
//http_request.overrideMimeType('text/xml');
//http_request.overrideMimeType('text/html');
}
}
if (!http_request) {
alert('Cannot create XMLHTTP instance');
return false;
}
switch(mode) {
case 1: http_request.onreadystatechange = alertContents1;break;
case 2: http_request.onreadystatechange = alertContents2;break;
default: break;
}
alert(parameters);
alert(url);
http_request.open('POST', form_url, true); // define the type of posting page POST/GET
http_request.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
http_request.setRequestHeader("Content-length", parameters.length);
http_request.setRequestHeader("Connection", "close");
http_request.send(parameters);
}
function constructString(obj) { // this constructs are poststr from the object
poststr="";
var i=1,name="";
for(j in obj) {
name = j;
val = encodeURIComponent(obj[j]);
poststr += (i == 1) ? name+"="+val : "&"+name+"="+val; //
i++;
}
return poststr;
}
// alertContents1 goes and gets any values we might need from our form.
// in this example we won't need this function
// the example below goes and gets form_token and creation_time then puts them in our poststr
function alertContents1() {
if (http_request.readyState == 4) {
if (http_request.status == 200) {
result = http_request.responseXML;
var form = result.getElementById("user-profile-form");
var action = form_url.replace(/\/user\/me\/edit/, form.getAttribute("action"));
var inputs = form.getElementsByTagName("input");
var inputdata = {}, inputname = "";
for(var i = 0; i < inputs.length; i++) {
inputname = inputs[i].name;
inputvalue = inputs[i].value;
inputdata[inputname] = inputvalue;
}
inputdata.signature = 'twilight';
inputdata.op = 'Save';
poststr = constructString(inputdata);
makePOSTRequest(action, poststr,2); // the 2 indicates calling the second function to send the data we got back
}
else {
alert('There was a problem with the request.');
}
}
}
function alertContents2() {
if (http_request.readyState == 4) {
if (http_request.status == 200) {
result = http_request.responseText;
//alert(result);
alert("You have successfully submitted to the form");
}
else {
alert('There was a problem with the request.');
}
}
}
function get(obj) {
poststr = "signature="+"twilight";//constructPoststr();
makePOSTRequest(form_url, poststr, 1); // we only call this function if we want to run alertContents1 as well
//makePOSTRequest(form_url, poststr, 2); // run only alertContents2
}
figured it out. The page I
figured it out. The page I was querying from had an error in it's markup so the code was exiting when it found the error.
Might want to check this
Might want to check this out...
http://drupal.org/project/ajax
thanks for the reply, but I
thanks for the reply, but I was submitting from outside the drupal system. Through a custom button coded for firefox. I got it though.