Using jQuery with validate plugin.I have to click twice on the submit button to get it working.First time I click on submit form = nothing happens. No error codeSecond time = form successfully submited in ajax.Here the code :
<div id="form">
<p><form id="contact_form" name="contact" method="post" action="verifForm.php">
<fieldset>
[code]....
I have some function to fire when I click a submit button. When I do that a form does not submit anything, it just fire my function etc. Here is my code:
$("#hm-submit-form").click(function () {
$("#nav ul li.products").delay(200).fadeIn();
$("#nav ul li.agel-tour").delay(600).fadeIn();
$("#nav ul li.compensation").delay(1000).fadeIn();
$("#thanks-box").delay(200).fadeIn();
$("#thanks-box").delay(1500).fadeOut();
$('#target').submit(); //this I guess should submit a form
I have a page built in ColdFusion where I have a text field in which user will enter a code (B12, for instanse). I am trying to make a lookup page using the value of this field. I am planning on passing the value of the field to the page where this value will be passed into a d/b query. User will not hit a button to send this page to the server yet. Is it possible to do it via JQ?
View 1 Replies View RelatedI'm trying to submit the form normally as well as by pressingCtrlEnter. Yepp, the same task.What I'm doing is the following:
var validator=$
"#myform"
.validate
[code]...
I have just started using jquery and am impressed so far.I am using jeditable to edit values in place, and submitting the values to a coldfusion page to update the values in the database.All this works fine.I am having a problem with the new (just submitted and updated) value being shown in the page.What would be the best way to achieve this? A request to the db to get the new value or just outputting the value that was just submitted?A sample of my code is below.
$(document).ready(function() {
//$('.edit').editable('http://www.example.com/save.php');
$('.edit_area').editable('http://element/admin/gallery/update-gallery.cfm', {
[code]....
What I want to do is replace a form with the submitted content. As an example, The form is submitted using the jquery.form plugin. The content is saved to the DB and a rendered comment is passed back to the client, and the success callback then replaces the form with the content. Simple enough. However, if server-side validation fails, I want to render the form, along with validation error messages, and insert that back into the client page, i.e. replace the first form with the second (which also contains the submitted content). Also straightforward, except that, on a truly successful submission, I'll need to do some other things on the page. If validation fails I only need to display the form with errors. I can think of two approaches. One is to search the returned data for some string (e.g., "<form") to decide if the other tasks are to be run.
The other is to send back the data encapsulated in a JSON object, eg:
{
'status': 'FAIL',
'data': '<form ...'
messages: '...'
}
Very intermittently, I am finding ajax requests submitted with jquery are being submitted twice, once with parameters, and once without parameters. The code looks something like this:
[Code]...
In my application,there are several autocomplte field is available.When I type "pank" and then it list out the several matching in the autocomplete field and select one and then hit the enter then it submitted the entire form, which will cause the error message. Because there are several more mandatory field are there, which i need to fill up.I was trying to restrict in this way, but i believe this is the wrong way.
View 1 Replies View RelatedI have email, password and some other fields, and I'm using $.post to send data.
In Ajax form submission, I don't want to submit a particular field. The serializeArray() method returns all the fields in the form. So, I tried something like this to prevent the password field being serialized.
This works great but the ajax submission doen't work.
I have a page which displays a list of resources in the database. For each of these resources I have a comment form. I am using jQuery form plugin to submit the form through ajax. After submitting I want to display the comment which was just submitted w/o reloading the page. But how do I know which form was submitted? I attach the resource ID as a GET variable to the action url of the form. If I can get the action url then I think I'm saved.
View 4 Replies View RelatedHow can I target a specific <div> to display the result of a form? I would like to be able to specify the target div on a per-form basis, and have the target independent of the <div> that contained the actual form.
So for example I may have a form in <div id="left"> <form>.....</form></div>, that I want to post the data to somepage.php, and load somepage.php into <div id="right">
I have a form which has a few submit buttons, all named SubmitButton so I can pass the clicked button's value through to the submission processor and then act according to which button is pressed. However on the jquery form submit even, I also want to check certain things before I fire the submission off. As the button isn't a "conventional" input I can't check the document object to see which one was clicked, so I need to check the form contents that are about to be submitted, can I do this from the submission event or will I have to bind to each button individually?
View 2 Replies View RelatedI use unordered list item (<ul>) to replace select type <select> (select) in my form. I have problem on bind the value after submit the form. Because <ul> is not a form element. May i know how should I overcome it?
What i intend to do is similar effect as google contact book: 1. User click ADD button, then <ul> and <textbox> will be appended into the form. 2. After form submitted, user can see what are the value he has entered earlier. I can bind the textbox, post the unorder list item with hidden element. but i have no clue how to bind the unordered list item..
Kindly refer to my code as below. It is working except cannot bind unordered list item... p/s: I am using PHP + JQuery.
<?php
print "<pre>";
print_r($_GET);
print_r($_POST);
[Code].....
I have a form I'm going to present to visitors. The goal is that when a visitor enters a Zip Code, for the form to compare that Zip Code submitted to Zip Codes that I'll hide in the form. I'm not sure if it's better to have the hidden zip code values as comma separated or just let them have their own individual hidden input fields. When the form is submitted, I need it to check and find out if the submitted zip code matches any of the hidden zip codes. If any of the hidden values match the submitted value, I want to be able to have a hidden div be shown, but if the form is submitted and there are no zip codes that match, show another div.
Something like this:
<form>
how to use the autocomplete jQuery plugin best to my needs. I'm looking to replace a select box with the Autocomplete plugin, but not quite sure how to accomplish this. In my select box, I have the option that is displayed is not the same as the data that is submitted.
ex: <option value="123">This is Displayed </option> However, with the autocomplete, since it is using a textbox, the values that are displayed, are submitted. Do I have to hack around using hidden fields to accomplish this type of behaviour?
I have a script that is supposed to: Make sure that the user selects a performance from a select field Make sure that the user hasn't selected a performance that contains the string "SOLD OUT" in its value Here is the code I am using:
$(document).ready(function() {
$('#paypal-purchase').submit(function() {
alert($('#paypal-purchase-date').val());
if($('#paypal-purchase-date').val() == 'Select Performance Date') {
[Code]......
[URL]
I have tried all the tutorial on the net which I can search, but all of them work without showing me the confirm message.
My submit form is simple as follow:
The user put their information in, the it will be valid using [url]
After the user submit the form, all the information will be sent over the internet to a mail.php file to process the sending.
The form disappear and a message confirmation appear.
However, all this process will not make the from refresh.
I cant really figure this out, the only thing i could see messing it up is the javascript:void(0) inside the anchor link but since there is a double click function and a click, it should only be one click.I put autoOpen which i think is also causing it, but i did that so theuser can open it, close it, and open it again so the delay "double click" is saying for the first time initialize and then the second click is opening?? if so how do i get around this?
HERE IS MY JQUERY
$(document).ready(function(){
$("#pro_edit_profile").hide();[code]....
HERE IS MY HTML FOR THE LINK
<a href="javascript:void(0);" id="pro_edit_pic_link" name="Change
Profile Picture">Change Profile Picture</a>
its possible to use jQuery to find and remove/replace an email address from a textarea input either before the form is submitted or after?
View 1 Replies View RelatedThere is a checkbox on a form page, when you check it, a div will become visible with some extra inputs needed for that situation. To simulate this click on the checkbox external, I use click(). With jquery files 1.4 and lower this works properly and like expected. When using jquery library 1.4.1 or higher, something does not seem to work properly. Sooooo, the code:
<script type="text/javascript" src="http://code.jquery.com/jquery-1.4.min.js"></script>
<!--<script type="text/javascript" src="http://code.jquery.com/jquery-1.6.1.min.js"></script>-->
<script type="text/javascript">[code]....
Somehow I managed to get this thing 'sort of' working with 1.4.1 or higher. But the code to accomplish this is absurd. See functioncheckTheBox4Plus() on line 8. It works :) but shouldn't.. Or is it the other way around?
I have a form,
It has multiple submit buttons (Save/Quit)
I want my form calidation to run when I save and not when I quit.
I still want to submit the form when I quit.
So how do I conditionally run the validation?
i have tried using submit() command to submit the form but the form does not get submitted. I wonder why.I am using lotus designer to develop the form:
var f= document.forms[0];
var answer = confirm('Do you want to referback?')
if(answer ){
[code]....
I have a form called "theForm". How can I tell if the user has submitted the form or hit the submit button?? Is it..
if(document.theForm.onsubmit()){
do something
}else{
}
Here is my Javascript / HTML:
javascript Code:
Original
- javascript Code
[code]....
I am relatively new to web programming (1.5 years) and this question just occurred to me.
I use jquery form validation to validate forms on my site. If my browser has javascript disabled, what is stopping someone from submitting crap through my form?
I tried disabling javascript and submitting a form on my site and it posted.
Is there a way to make it so if the user does not have javascript enabled the form cannot post?
I'm trying to submit form data to mysql database. The POST form action calls a javacript/ajax httpRequest. My php script has a redirect at the end of it on completion of data submission. My problem is after submitting the form, the redirection works, but no data is entered into database. I'm wondering if there is a problem with my httpRequest.
[Code]...