JQuery :: Get Response From Php Script Called With Form Plugin?
Apr 20, 2010How do i get response from php script called with form plugin?
View 1 RepliesHow do i get response from php script called with form plugin?
View 1 RepliesI'm having troubles with an existing J2EE application (which uses Dojo) and in which I'm gradually introducing jQuery. The specific issue is with the malsup Form plugin and the JSON returned from a form submission: the error callback is always called, regardless of what happens on the server side, and the error is always "parsererror". I'm using jQuery 1.4.2 and the 2.45 version of the malsup Form plugin.
For example, given the following code:
$(document).ready(function() {
var options = {
dataType: 'json',[code]....
the server receives the submission and handles it without errors, then in the browser I always get the same alert from the processAddressEditSubmitError() function, with an "Invalid JSON" message:
pStatus=parsererror
pErrorText=Invalid JSON: {"nickname":"trytrez","success":"success"}
However, the JSON looks OK for me (and http:[url]...agrees that it is valid).
I am using this plugin in order to upload pdf or excel files. I have not problem with Firefox but only with IE. Unfortunately, as I read on the forum, there is an issue concerning getting the response from the iframe, even if I can get the uploaded file. I need that functionality in order to get error messages from the server that will be displayed on the user interface. I know that it will be solved in a future version of the plugin but I wonder if there is a workaround. I made several attempts and I arrived at the point where a cross domain issue stopped me.
View 1 Replies View RelatedDoes anyone know how or if this its possible to get the response headers using the jQuery Form Plugin? I've got a success function which needs to get access to a response header I'm setting.
View 3 Replies View RelatedI'm using Malsup's jQuery Form Plugin and trying to return a json object:
$('form#new_form').live('submit', function(e) {
//e.preventDefault(); // <-- important
$(this).ajaxSubmit({[code]....
Currently using: version: 2.47 (04-SEP-2010)
If I have iframe: true, my JSON response back is always:
<head></head><body>{"WEBSITE_URL":"test","ASSETTYPE_ID":0.0,"ASSET_RESULT":{"ASSET_ID":9,"ERROR":false},"ASSETTYPE":"link","NAME":"test","DESCRIPTION":"test"}</body>
Was this done for a specific reason (note the Head/Body tags)? Is there a safe way of parsing this JSON request out? Am I doing something wrong?
I have a xhtml jsf page that load anothers xhtml jsf pages that have, each page, 1 table with tablesorter enabled, when i load the first xhrml into the DIV the tablesorter works ok, when i load the next page the tablesorter od the page loaded before stops to work. In fact,after debugged the table sorter code, I found that the header.click() of the tablesorter loaded before is being called twice.
View 2 Replies View RelatedI'm initializing a lightbox type plugin after an ajax response. The plugin is usually initialized with .ready, but I've read on this forum that .ready won't work after page load or with dynamic loading. Most posts say to initialize right after the ajax response... I've done this and it works for the first ajax load, but not for the subsequent ajax loads. It should trigger the same code every time ajax is called since it goes through the same function, but it doesn't.
Project Info: This project seems pretty simple... There are links that load content into a DIV via ajax... the content is made up of thumbnails that should launch lightbox...Is there a good way of dynamically initializing plugins after ajax load?
I'm using the uploadify jQuery plugin from here: [url] & am having some problems.
Everything is working correctly but the uploadifyUpload() function doesn't appear to be triggering. It is not accessing the add_list.php script whatsoever.
Here is my code from my header..
Code:
It correctly shows an alert box with "Correct" so I know it's getting in there to run it, but it just isn't running.
Shouldn't need to see this, but here is the errorCheck() function if interested..
Code:
Lastly this is the HTML code that calls it.. stripped out a lot of uneeded code for clarity.
Code:
m relatively new to Jquery and have come accross these two plugins.Having looked at the ajax examples offered for the form plugin i wasintruiged to find out how i could go about validating the form usingthe formvalidate plugin during the beforeSubmit callback.Ive seen that you can validate the ajaxform as shown in the followingxample.malsup.com/jquery/form/#code-samplesHowever i'd like to use the formvalidate as it offers alot more....
View 1 Replies View RelatedI have an ajax post which returns a large html response. It is getting truncated at 98784 characters everytime. Is there a limit to a response size or a way around this?
Using: IE7
jQuery 1.4.2
jQuery UI 1.8.1
Here is my ajax call:
I'm having an issue using the jquery form plug-in. The form gets submitted just fine I can see the response in Firebug
[Code]...
I have ajax request with a success function. The data I'm getting back are an entire <html> ..</html> page. Is there a way to retrieve from those data only a given DOM element by its id ?
$.ajax({
type: "POST",
url: requestURL,
data: formData,
[Code].....
Having looked at the ajax examples offered for the form plugin i was intruiged to find out how i could go about validating the form using the formvalidate plugin during the beforeSubmit callback. Ive seen that you can validate the ajaxform as shown in the following example. [URL] However i'd like to use the formvalidate as it offers alot more..
View 1 Replies View RelatedI am trying to create a form that writes text to an HTML canvas when submitted. Eventually, the function that writes the text will be more complex. The problem is the text only appears briefly, because the function is only called once when the form is submitted. I want the function to be called continuously after the form is submitted.How do I do this? I have had very little experience with JS.A lame (failed) attempt...
<html>
<head>
</head>[code]......
I have designed a form as below
<form action="https://checkout.google.com/api/checkout/v2/checkoutForm/Merchant/XXXXXXXXXXXX" id="BB_BuyButtonForm" method="post" name="BB_BuyButtonForm" target="_top" >
<input name="item_name_1" type="hidden" value="Test Product"/>
<input name="item_description_1" type="hidden" value="Test Product of my cart"/>
[Code].....
On hitting the button one new window is getting opened with response URL from 3rd party(in this case Google Checkout). But for me i need to open this in a inline-popup (lightBox, ColorBox etc...) rather than new window. In the form definition, i have replaced mymerchant_idwith 'XXXXXXXXXX' for privacy purpose
I want to submit my problem. I use the plug-in "Validation" to check and send my "form", and thus far everything is working correctly, just do not know how to retrieve information from the page to which I have submitted my form, such as errors or other info.
View 2 Replies View RelatedWe're using Malsup's form plugin v2.82 and jQuery core 1.6.2. We're having issues in IE7 where, on a form with file upload, form fields are being dropped before the form is submitted to the server.We tracked it down to the `fileUpload` function, specifically line 196 where each field in the form data has `.attr('disabled',false)` run against it. If we change this to`.prop('disabled',false)` it all works fine.I don't see a clearly marked place to submit bug reports for the form plugin, and wanted to verify that others see this as well,
View 6 Replies View RelatedI'm trying to get an upload popup working with the jQuery form plugin[URL].. When I click a link I load a form html from the server and add it to a container div by setting the div's html attribute. I then attach a submit handler to the form so I can call the ajaxSubmit function of the form plugin.
[Code]...
Malsup's most excellent and comprehensive Form plugin has me completely stuck on just one thing.Take a look at this: http:[url]....At the bottom are a variety of submit buttons, and when you click one, it knows which one has been clicked.And I've been through the js and the source and the examples and I can't figure out how the bleep it's done!I'll tell you why I'm asking, then perhaps you can probably tell me I'm doing it wrong anyway!Let's say a blind person logs in, and want to edit their presets.I don't want the form to be too complex or clever or ajaxy, as screenreaders don't like that, so it just iterates through as many presets as they have, and populates a form with edit boxes.But there's no point "pushing back" 29 unchanged items just to edit one row.
So my idea was I'd just "fieldSerialize()" the details of the row that was currently being edited and submit that to my little php routine that updates the db. Then they can do a refresh just to hear the list again.The js looks like:
$(document).ready(function() {
$('#myForm').ajaxForm(function() {
var queryString = $('#myForm').formSerialize(); [code]....
All works fine like that. But if I change line 3 to: var queryString = $('#myForm :button').fieldSerialize(); it doesn't work. I've also tried:
'#myForm :button'
'#myForm .button'
':button'[code]....
Maybe I should just generate as many separate forms as there are presets, but then I'm going to need as many ready(function)'s as there are rows, which is going to be very messy.
I'm using the Form Plugin to allow a user to upload a photo.The user clicks on a link in the page and I display a dialog collecting the file and description.It works great the first time, but when they do it again the description and a hidden field is not passed n the form.How can I get the form to work more than once without reloading the page?I call this on the click of the link on the page.
var options = {target: '#divToUpdate', success: PhotoUploaded};
$('#photoForm').ajaxForm(options);
I have tried the jquery form plugin as in the examples and it worked pretty well.
Then I tried to handle the following form:
<form
id
=
"loginForm"
[Code].....
I needed access to the form that's being submitted in the options.success() function (in order to hide the form after successful submission). I ended up modifying line 222 in jquery.form.js. I added $form as the third argument for all callbacks, like so:
options.success = function(data, status) {
for (var i=0, max=callbacks.length; i < max; i++)
callbacks[i](data, status, $form); // added $form as third argument.
};
Is there a better way of accessing the form from the callbacks or would the plugin developer(s) consider adding this or another similar modification to the code?
I have my form working great as long as my submit button is contained within the form tags. But the design calls for the submit button to be outside of the form. Sample code and diagrams are below.
<form>
my form here
</form>
<div> </div>
<div> Submit button </div>
[Code].....
I am currently trying to validate a form before sending it with the jQuery Form Plugin.I can get them working but it is always one or the other, I can't get them both working.
[Code]...
I want to get the solution for a null xml response at the moment if there is no error i made a work around so the xml returns a fullstop which shows up as an error but in white. I would like to know if there is a way to keep the elementByiD as null when the xml is null as it was returning the error object required is there a better way of doing this????
[Code]...