I see it working in the demos I found this documentation
You can avoid having to duplicate the plugin settings by modifying the defaults. Use $.validator.setDefaults({…}) to override multiple settings at once.
But I do not understand what I have to do to make this work
I've a page that has 2 forms. If the user clicks the bottom form the validations appear and everything goes red but the Top form is a required field but it doesn't as it's a separate form. How can I validation form0 and form1 together to show to the user the required fields. When the user clicks form0 it goes over to a pagedLIst, the user returns and textfield gets populated with the product that as selected. So when they hit form1, that value gets passed through as a hiddenfield..
<form id="form0">
...text box input to search for your product, returns to this page populated the found product
<input class="button" id="FindProduct" type="submit" value="Find Product" name="action"/>
[Code].....
I know that this must be quite trivial to most of you, but my skills
in JavaScript are weak.
I am looking for a way to validate all forms throughout a website
through one validation script.
How do I determine the form's name (or id) value to implement
conditional validation?
i.e. if (form1) { validate like this }; if (form2) { validate like
this };
I will have 1 html page which uses javascript it will validate the form so the information inputed is correct...then on submit it will load another page with form ect...however the second page depends on what was selected on the first page say a gender radiobutton was on the first page and they selected male then on the second page a male figure/picture would be displayed...now I can do this all on 1 page however when trying to do it on 2 different pages the second page can't read whats going on/has happened on the first page...
View 3 Replies View RelatedI have a form with multiple submit buttons.
<form action="" id="myForm" method="post">
<fieldset>
<!-- various input fields -->
<button type="submit" name="Exit">Exit</button>
<button type="submit" name="SaveExit">Save & Exit</button>
<button type="submit" name="Save">Save</button>
</fieldset>
</form>
When the SaveExit or Save buttons are clicked, the form is to be validated. When the Exit button is clicked, the form is NOT to be validated. How can I do this with the Validate Plugin 1.7 from [URL]. Is this possible?
I'm struggling with a script for an e-commerce site.When the user clicks the add to cart button they are currently taken to a PHP shopping cart page, but the client wants them to stay on the same page after they press the button and get notified that an item has been added to the cart.The add to cart button is a form submit button, because there is also a quantity field for each item. I would have cracked this by now but there are multiple add to cart buttons on each page and when I submit the form only the data from the first form in the page is sent to the cart page.Here's a dummy version the JavaScript:
Code:
$(document).ready(function(){
$('.subbtn').click(function(){
$('<p id="add">Added To Cart</p>').insertAfter(this).fadeOut(3000);[code].....
At the moment I've got it set so that AJAX returns information to the page (just so I know it works), but this won't happen on the actual page.So, as far as I can tell, I need a way of POSTING $(this).parent(); $('[name=quantity]').val(), but I can't seem to find the correct syntax.
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 RelatedThe page I'm working on has a week's worth of data blocks like this - pseudo code & markup:
<form 1>
field a 1
field b 1
submit 1
response 1
</form 1>
<form 2>
field a 2
field b 2
submit 2
response 2
</form 2>
and so on for either 7 or 14 days.
I'm trying to conceptualize an prototype based js function that can handle all the forms to submit the data to the backend.
What I have now, that works is like the following
Event.observe(window, 'load', ajax_init, false);
function ajax_init () {
Event.observe('form_1', 'submit', send_form_1);
Event.observe('form_2', 'submit', send_form_2);
}
function send_form_1 (e) {
$('response_1').innerHTML = 'doing the deed for 1!'
var myAjax = new Ajax.Updater('response_1', 'ajax_server.php', {method: 'post', parameters: Form.serialize(this)});
Event.stop(e);
}
function send_form_2 (e) {
$('response_2').innerHTML = 'doing the deed for 2!'
var myAjax = new Ajax.Updater('response_2', 'ajax_server.php', {method: 'post', parameters: Form.serialize(this)});
Event.stop(e);
}
But, in my mind, there has to be a way to streamline this so that one function 'send_form' can do the magic rather than having multiple iterations of it. I was looking at the bind functions in prototype, but perhaps it was the late night and lack of coffee, but I wasn't seeing the application of bindAsEventListener in this context.
I have a php page with java script embed in it, i have 3 forms in a single page ... namely form1, form2, form3 . form 1 and form2 have a text box , these text boxes get input from users. Now my task is to get the value of these 2 tex boxes in form3 when the user click the submit button, so that i pass the values in those text boxes to next page for calculation.
I tried the following code in form3 but it result in vain :
I'm using the validation plugin [url]
I must be missing something, is there not an easy way to validate inline?
For example, I have a required text input that is given focus on page load, I would like for it to throw an error if the user moves to the next field w/o entering any data.
Right now, it only checks for errors on 'submit'.
I am using the plugin jquery.innerfade to make a small slideshow, but I am having trouble using it for two. I want two different slideshows on the one page, and in different divs. I can't seem to get it to work.
Code:
<script type="text/javascript" src="js/jquery.js"></script>
<script type="text/javascript" src="js/jquery.innerfade.js"></script>
<script type="text/javascript">
$(document).ready(
[Code].....
I have a form (form1) to register with the fields: user, email, password being validated normally.
Have a second form (Form2) to the user update the data registered in form1, and the fields user, email, password is already populated with data from the database, the email field I use the remote validation, and as the field already has been completed and read always says that this email already exists, how do I create a rule or method so that when the field is already filled it not do so validating and validate if it is filled with a new email. Email2 have a hidden field in order to do a test type.
I have a page with 2 product forms to add to shopping cart each form has the same fields.
when someone changes one of the dropdown boxes depending on the size it is to change the price to a certain value. This works ok whan you only have one form but now that there is two forms. When you change the second form dropdown box it changes the price on the first product.
OnChange='if (this.value<=16)ChangeHiddenPrice(90);'
I suspect it has to do with how i reference the form field
function ChangeHiddenPrice(newprice) {
document.getElementById('PRICE').value=newprice;
}
I have two scrolling RSS feeds at [URL] at the bottom of the page.
I've created these by calling a php document containing the script and the feed.
I can't get any more than one to work. I've tried it a hundred different ways and no joy.
I'm working under certain constraints wherein,at a certain point outside my direct control,validate is called with no arguments, but after that I want to set a custom validation function on a form field. I figure there's got to be a way to manipulate validate's internal data structure to add the function, but I don't have a clue as to how.
View 1 Replies View RelatedI have a section of a form containing elements that are not visible, currently just in a hidden div. Some of these elements have validation on them so when the form is submitted it fails on this validation.What I am after is a way to stop the validation firing for any hidden elements. I have tried setting them to 'display: none' and 'visibility: hidden' but this does not have any affect.
View 5 Replies View RelatedI am looking for a plugin to paginate a multiple page form. I would like to decouple validation and navigation forward and back.
What plugins or demos can you recommend for pagination of forms?
I have a problem in form submit.
I have multiple forms in my page with different name, instead of submit button i have images, on click of image i am submitting the form. but while submitting its giving Object doesn't support this property or method error:
I have a form with 12 fields. This form submits the information for registering a student for courses that we offer. Currently, it only handles one registration at a time and is kind of clunky.I want to present the user the capability to register more than one person at a time. Essentially adding multiple users' info at once before submitting the form.I also want to have an area above the form that shows a list of people that they enter in a select,which will later be modified to have checkboxes so they may remove them, if needed.
View 2 Replies View RelatedI need to localize my validation plugin in general. Hence I thought I have to override the default code with this one:
But it's not working. Another problem is, that I can't access input names like this > name="xyz[1]" Is there a way to do so?
Using Jquery Cycle Plugin i have multiple containers on the same page. they have a next and prev button. the problem is, when i click the next button it will cause the containers on the pageadvance simultaneously, same goes for the previous button. it works fine when i click just on the images (they will go to the next image) but not with the prev. and next. button. i know i have to add some kind of var. to it. and i tried it but with no luck.
Here is my code.
$('div[id^=slide]').each(function(index) {
$('#slide'+ (index+1)).cycle({
fx: 'fade',
speed: 'fast',
[Code].....
im using the validate plugin with the meta data plugin
jQuery(document).ready(
function(){
jQuery("#com-createForm").validate({ meta: "rules" } );
});
and then in the html i have for example
[Code]...
I have a page that relies on AJAX for content manipulation. Once a person selects a radio button, a request is made through $.getJSON() to get the replacement content. This works in all browsers. Once the new content is loaded, there are 3 more calls to $.getJSON() to change more content (on other tabs in jQuery UI). This works in all browsers except for IE.
I understand this is a Microsoft problem, and not a jQuery problem, but I am wondering if there is a way to manipulate the jQuery calls to get them to work in IE. I did try replacing $.getJSON with $.ajax, and even set async to false. None of those three methods worked in IE. I noticed the problem when my loading div was still displayed in IE. Using Firebug Lite, I saw that only one request was made, rather than four.
I'm a javascript rookie trying to figure out a way to get bookmarkable URLS for the images on a single page with multiple slideshows. I must admit I don't really understand the first part of the code on the Cycle demo page for this, which only works for a page with one slideshow:
While I'm shopping for a little guidance, it would also be really cool if the hash tags were semantic somehow with respect to each slideshow e.g.:
[url]
[url]
[url]
I am rather new to JQuery and would like to use it with php. I ran into problems when my html is not static, I generate php forms on the fly using my php script and JQuery just doesn't seem to validate those forms. When put the same form in static html file JQuery works like a charm. What could be wrong? Does JQuery need a page to be static? Right after I load my php page firebug reports that $(document).ready(function() { is not a function and JQuery is not defined. I have loaded the necessary JQuery libraries.
View 3 Replies View RelatedI am trying to change each forms .submit function like so (below) but each submit button gets the function of the last iteration. I want each form to have a different submit function without using onclick events.
var x = document.getElementsByTagName("form");
for(i=0; i < x.length; i++){
var ele = x[i].elements;[code].....