I 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].....
I would like to allow the user to be able to use the Enter key in lieu of the submit button on a page that has several forms.I'm assuming that toaccomplish that I need to detect which form's elements are being edited (have focus?) and programmatically fire the submit button's action for that form.How do I do something like that?
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f u cn rd ths u cn gt a gd jb n prgrmmng
I am setting up a text search with 3 search buttons to allow 3 different searches from the one text box. Each search has different values for the 2 hidden elements. So far I've got the 3 submit buttons working with the below code but I can't figure how to get the hidden values to be inserted. For each of the 3 different submits I need to tell javascript what the 2 hidden values are.
<SCRIPT language="JavaScript">
function OnSubmitForm()
{
if(document.pressed == 'questions')
{
[Code]....
If I disable JS is running everything as I need. (but it isn't my objective)
<form id='RequestData' action='request.php' method='post'>
<button type='submit' name='par[1]' value='V1'>
<button type='submit' name='par[2]' value='V2'>
[code]....
Script work, but send wrong data, always send to request par[3]='V9' how I can do to send data' from buttons which I click ?
Myobjective it data:
par => array(
[1] => 'V1'
)
only one value of array PAR
I'm trying to use a single submit button to submit 3 forms on a single page. 1 form I use and process, the other 2 are sent to other sites.
Everything works great except in Google Chrome (and Safari) due to some restriction/bug in WebKit.
Code:
This works perfectly fine in IE and FireFox, but in Chrome - as it sits, it will submit the first 2 forms only. Yes, I know the return false prevents anything else, I'm getting to that.
With the return false in, it will submit the 2 forms. Removing return false only processes the form named Test1 and not the other 2.
I've shown code for attaching to the click event hoping that I could sneak my submits in on that, then let Chrome carry through to the submit function, but that doesn't work either.
I have built out a fairly rhobust app using jquery and jquery-ui all was going well until we started to bring different screens together inside a tabbed interface. Seems that now when we execute our validation code on one form its checking the fields of all other forms on the page.All of my forms do have unique ID's and all elements within all forms have unique ID's.My understanding of using this.find inside of the submit event was that it should only find the forms elements am I wrong? Method that performs the actual validation:
function validateForm(event){
var allOk = true;
$(this).find("*[validation]").each(function(){[code]....
How can I use jQuery to submit multiple forms with different types, values, and actions? For example:
Code:
<form name="form1" method="get" action="submit1.php">
<input name="value1" type="text" value="">
</form>[code]....
Can I start the jQuery process from within a function when clicking on the submit button?
I have a form that has 3 submit buttons with 2 input fields. The first input field is a search field that allows the users to search against the database for certain names that are found and not found.[code]...
View 1 Replies View RelatedI have a form with several submit buttons. I've used $(#myform).ajaxForm(options) method to make the form submittable with ajax. the problem is that no matter what button I click the form data is the same and <input type="submit"> element is not included into the form data, as if it were not a "successul control". so at this point form doesn't work as expected. is there a way to convey information of which button was clicked with the form plugin?
View 4 Replies View RelatedI need to submit two forms using a single submit button. What's ur suggestion. I found that JavaScript is the only way to go here, right?
View 1 Replies View RelatedHow can one submit multiple forms at a time. At the same time the values must get Forwarded another page.
View 4 Replies View RelatedI have several forms on a page, one for each "product". Each form is essentially a row with product information. I have a "Add to Cart" button. Now, I want this to work so that when you add a product to the cart, it does it without refreshing the page (AJAX?)
I found this [url] and it works to an extent... it updates the DIV for the first product only (and understandably)
I am trying to add an additional submit button to a form that has a separate action. This is for a shopping cart. The button I am trying to add is a "Preview" button. Which will have a different action then the "Add to Cart" button. Right now it works but after the "Preview" button is clicked the "Add to Cart" button will use the action that the "Preview" button is using. This is the script I am using for the "Preview" button.
var productPreviewForm = new VarienForm('product_addtocart_form');
productPreviewForm.submit = function() {
if(this.validator.validate()) {[code]....
I have a form with several fields.
Each field can be modified:
- individually (one submit button next to the field)
- globally (a submit button at the end of the form).
Before submitting the form, a validation needs to be performed.... so I added: onSubmit="return isValid()".
If the validation does not pass, the form needs to be shown with the data previously entered by the user.
Now, I haven't even come to the previous problem I have mentioned..... my problem is that if a field is empty, the function isValid() recognize it (message sent) .... but the reDirect function is performed!
I am using jquery with the cookie plugin and I have multiple image buttons that can hide/show multiple elements. My question is how can I add a cookie to this code to remember whether each separate element is opened or closed?
The code,
$(document).ready(function() {
// choose text for the show/hide link - can contain HTML (e.g. an image)
var showText='<div class="expanddown"></div>';
var hideText='<div class="expandup"></div>';
// initialise the visibility check
var is_visible = false;
// append show/hide links to the element directly preceding the element with a class of "toggle"
$('.toggle').prev().append('<a href="#" class="togglelink">'+hideText+'</a>');
// capture clicks on the toggle links
$('a.togglelink').click(function() {
// switch visibility
is_visible = !is_visible;
// change the link depending on whether the element is shown or hidden
$(this).html( (!is_visible) ? hideText : showText);
// toggle the display - uncomment the next line for a basic "accordion" style
//$('.toggle').hide();$('a.toggleLink').html(showText);
$(this).parent().next('.toggle').slideToggle('fast');
// return false so any link destination is not followed
return false;
});
});
HTML,
<a class="togglelink" href="#"></a>
<div class="toggle">
Content
</div>
I have a complicated page with several submit buttons. I don't want to
create multiple forms because much of the input is shared and the code
is getting very ugly. However I would like to determine which values
will be submited.
Here is a simplified example that may better illustrate my question.
Or maybe there is a better approach...
<SCRIPT language="JavaScript">
function submitform1()
{
document.myform.submit();
}
function submitform2()
{
// logic to submit only name parameter
document.myform.submit();
}
function submitform3()
{
// logic to submit only address parameter
document.myform.submit();
}
</SCRIPT>
<form name="myform" action="#">
<TABLE>
<TR><TD>Name</TD><TD><input type='text' name='Name'></TD></TR>
<TR><TD>Address</TD><TD><input type='text' name='Address'></TD></TR>
<TR><TD>Phone</TD><TD><input type='text' name='Phone'></TD></TR>
<TR><TD></TD><td><input type="button" onClick="submitform1()"
value="submit all variables"> </TD></TR>
<TR><TD></TD><td><input type="button" onClick="submitform2()"
value="submit just name"> </TD></TR>
<TR><TD></TD><td><input type="button" onClick="submitform3()"
value="submit just address"> </TD></TR>
</TABLE>
</form>
I'm trying to search for the correct code to make my form work. I have 3 select boxes - one with 2 options, one with 8 options, and the last with 2 options... All of this adds up to 32 different url paths. Can anyone tell me how to get this done?This is what I have so far: (and yea, I know I suck... I honestly have absolutely NO clue)
<html>
<head>
<script>
[code]....
I have a form with multiple fieldsets which are visible conditionally. There are three submit buttons "Abandon", "Save" and "Save & Continue". Each button should validate specific controls of the form and submit it. I tried setting "onsubmit: false" and checking for "$('#myForm').valid ()" on click of these buttons., but that validates all controls of the form.
View 1 Replies View Relatedhow to assign multiple action with multiple submit button in the form.
View 2 Replies View RelatedI'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'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 RelatedIt doesn't appear to be possible to validate multiple forms on a pages that only has a single <form> tag, using the JQuery Validation Plugin offered by Jörn Zaefferer. The reason I need to do this is because my pages are created with ASP.NET whose architecture generally calls for a single page-wide <form> tag. In my designs, I'd like to submit my data using a custom ajax call to a web-method, and link it to a click event on a button.
Below is an example:
<!
DOCTYPE
html
PUBLIC
"-//W3C//DTDXHTML1.0Transitional//EN" "[URL]" >
<
html
xmlns
=
"[URL]"
> .....
I'm not sure that I fully understand the documentation for the Validation plugin, but it doesn't seem like this is possible, as the plugin seems to need a <form> for each form.
[URL] that is the code in all its basic form (just stripped out other html). I am trying to modify the form actions of the 3 forms when i click on the tabs at the top of the page. but only the action of the delete form (id=form-horse-delete) gets modified. the other 2 forms do not change. i am not sure what the problem is. in IE8, if i have the jquery modifying the action of the 2 forms (add/delete), it gives a JS error. if i take out the add/delete in jquery, there is no JS error.
View 3 Replies View RelatedI 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
Is it possible to post all (or more of one) forms simultaneosly?
document.forms.submit?