I need some assistance with sending JSON to the server. When I make this post using $('#trfFilterForm').serialize() I can view that URL-encoded string on my server (cherrypy). When I make the post using $('#trfFilterForm').serializeArray() the server responds with status of 400 Bad Request (I'm not even touching the JSON on the server side at this point). Is there something else I need to be doing to the JSON data before I post it?
Code:
I noticed that serializeArray() method does only include those elements that does not have the 'disabled' attribute... I wonder if it's intentional or am I doing something wrong...?
View 1 Replies View RelatedI want to run an external function outside the post.
This is what I have currently.
On success of the post I want to run the setGrandTotal(); function which will do some calculating for me.
How can I post array of controls through &.post() in jQuery??I have some of checkboxs in my page and declare them like this
<input name="mycheckbox[]" "value="myvalue" type="checkbox">
I can get it works if I using <form> tag with an action="post" and a submit button to post to another php page by using this statement to get all controls in receiver's page _GET['mycheckbox'];I have tried
$.post('ajax/test.php", {checkbox: $
(
'input[name="mycheckbox"]'
)});
[code]....
I just started using jQuery, but i can't get it working. On the index.php page I want a search form, that post's to search.php. Following next, I want that the html of search.php (which will only be a table with the results), is show into the 'results' div in the index.php.
This is the code im using:
<script type="text/javascript">
/* attach a submit handler to the form */
$(document).ready(function(){
alert("Ok - 1");
[Code].....
The alert's are for debugging, but none of them show's up.
[Code]....
how do i cause the form to post to another page? Am I doing it correct? I want this to happen ONLY if validation passes.
This form field for a PayPal transaction will be produced during a user session. how to automatically submit it with "live" and $.post?
<?php
SCRIPT
print "<form id="paynowForm" action="https://www.paypal.com/cgi-bin/webscr" method="post">
[Code].....
i need to post data from a form to a page on another domain, without leaving the current page.I am using ASP as a server language.
View 9 Replies View RelatedAm calling Webservice in one ajax post, In the success funtion am calling another another method in same webservice through another ajax post method. First ajax post is getting called and returning the string from the webservice method but the inner ajax call is not getting called. I have placed the code here.
[Code]...
have a problem when using get or post from subdomain.I have a file included in header: functions.js. in this file a have a function similar to: function popup(thevar,json){$.get([URL]...{thevar:thevar,json:json},function(data){alert(data);});} It works fine, but when i try to use this function from subdomain.domain.com it fails with no error. Is there a way to resolve this so that i can call the function from any subomain?
View 1 Replies View Relatedhow to send data to another external php page from current php page and also get the html code from that external php page.
View 4 Replies View RelatedI am trying to make an ajax POST request but its not working. Code below:
var _HOMEDIR = "http://localhost/personal/index.php/home/";
$(document).ready(function() {
$("form").submit(function() {
var str = $("form").serialize();
[Code]....
I want to send data to a database by a classic asp page. This page returns xml which I want to parse on the client.
[Code]...
I have this short code:
// JS code
$("form#submit").submit(function() {
var fname = $('#fname').attr('value');
[code].....
<script type="text/javascript">
In the above code which I understand to be jQuery I understand what $('#search_results') is doing but I don't understand at all the purpose of "$.post". Is there not supposed to be something before "$.post"? What is it doing? If search_results.php is being called then what is the purpose of { f : "search", zipcode : zip }. Why is
function(data) {
I really am lost and could use a good explanation of what is going on here.
why to use the .post instead of .ajax?
View 2 Replies View RelatedI know that there can be a 3rd argument for Jquery's $post. I want to figure out what the best solution is, for a function, that can be used so we can guarantee that no other jquery functions run until the $post action is completed. I use $post to hit 'vote.php' to submit a vote but I want it to complete this process, and be able to confirm it (Without an annoying alert) so it can properly continue with the process of displaying vote results to the site visitor. I just am not sure what I should do. What is the best idea, just create a function that displays a 'please wait' image that loads for a few seconds and then continue?
View 10 Replies View RelatedI posted this problem in the old mailing list, but I thought Id try here with a larger set of eyes.
My issue seems to concern webkit browsers. I am sending information from a form with a jQuery click function. I am loading the response from one form that is a DIV called rightContent to the same DIV and passing the
[Code]...
IE does not seem to make the post using the following code:
$(function(){
$("#update").click(function(ev) { var target = $(ev.target); if (!target.hasClass("add_btn")) { return false ; }
var li = target.parents("li"); var _title = li.attr("data-title");
[Code]......
I am building a web app that uses alot a jquery including a function that reaches out to a webservice using $.ajax to grab data and return to tdata to the web app. The function that makes the call is outside the ready block but is called in the ready block to make it run on the load. After the load I use a timer to keep calling the function to keep things up to date. But after the first run the function does not seam to work. This might sound wierd, but the ajax function appears to run after the initial go, but it always returns the original request, it does not make new request to the server. I use fiddler and watch what going on and after the initial run method it doesnt make any other calls even though I see the ajax method being stepped through using debugging software and see the AJAX call being attempted and returned successfully.
var REQUESTURL = '/citizenQuestionWCF.svc/';
var prOptions = '';
$(document).ready(function () {
getQuestions();
[Code].....
So what I thought was going to be really turns out to not be working for some reason. I'm setting up a shell for my new site. So I'm just trying to get some of my pages linked together for testing the site flow and feel. I want to do this with javascript (jquery) because when I start passing session data that's how I'll be handling this.
So I have this simple line in my js code...
$('#submit').click(function(){
$.post("plans.html");
[code]....
I'm a total JS/jQuery newbie, and I'm having a problem using jQuery's load() to POST data to a remote PHP script and rewrite a DIV with whatever the server sent back.
Using a proxy in between, I can see that the JS script isn't sending anything, so I guess there's something wrong in the JS code I send to the browser [code]...
I am very new to Jquery so sorry for the simple question. At the moment I am trying to create a form that posts to a cgi script then returns the results in a div on the page. This is working fine. The problem is that when the results are shown and I go to submit another query the next set of results appear under the previous set, and so on. I would like to remove all previous results as soon as an new search is submitted.While I can get this to work with the code below it willbrieflyshow all the previous results until the query iscomplete.
[Code]...
I would like to delay my jquery .post() call for 5 seconds like: jQuery.delay(5000).post( ajaxurl, data, function(response){ alert(response); } ); but it is not working. the error msg tells that the object do not support that method jQuery().delay(5000) or jQuery('body').delay(5000) is too not working.
View 8 Replies View Related" $.post(Drupal.settings.newURL.admincharacteristics+ 'display/' + encodeURIComponent(a) "In tha above code, I want to have variable "a" to be an array?
View 5 Replies View RelatedIf I have this:
$.post('/admin/duration', $str, function(data){
var jsonData = eval('(' + data + ')');
return jsonData.TIME;
});
How do I get the return value from the function to use afterwards? I know this must be simple. I'm just not seeing it.