- 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 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.
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 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') {
I can get the submit button to work and the above button to work.The issue I am having is that I have jquery validation in the form and whenever I hit the SAVE button it wants to validate the form (which I dont want it to as I want it to save at any point in the form). Does anyone know of a quick way to bypass validation code in jquery. Say with using an ajax call and form submit or someway to say if the SAVE button is clicked ignore the validation plugin.
I have a form, that uses validation with jquery, it contains 2 submit-buttons: one is 'back' the other is 'forward'
How it should be: By clicking the 'forward'-submit-button, jquery should validate, if clicking the 'back'-submit-button, jquery should ignore validation
But the problem is: jquery validates at each submit
Now my question: How can I get it, that there is no Validation, when I click on 'back'-submit-button, or in other words, that jquery can differentiate on whitch submit-button it should validate?
I'm validating a form that has two buttons ("Next" and "Back"). The call to the script is currently in the form's onSubmit() handler. Upon pressing either button, the script runs. This is understandable, since either button press is considered a submit. Is there a way to test which button was pressed to determine whether or not to run the script?
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 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]...
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>
I 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?
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? ____________________________________ f u cn rd ths u cn gt a gd jb n prgrmmng
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]....
I have a problem where if a form submission (set up to submit via AJAX) fails validation, the next time the form is submitted, it doubles the number of post requests - which is definitely not what I want to happen. I'm using the jQuery ValidationEngine plugin to submit forms and bind validation messages to my fields. This is my code below. I think my problem is that I need to unbind from the validationEngine plugin when the form fails, but I can't figure out how to do this.
On my form I have 3 submit buttons which handle different things.I am looking for a way to stop or continue form execution with a confirm box on the third submit button and the third only.I can't use onsubmit because that will trigger on all three buttons.
I want my radio buttons to become submit buttons as well. So when a user clicks on a radio button it submits the action and refreshes the page accordingly. As of now using only using onclick="this.form.submit()" the page only refreshes with no change. Here is a copy of the entire form. It is a custom shipping options form (I did not create it).
I'm using the Validation plugin for JQuery and was wondering if there was a function to submit the form without causing it to validate the form. I have a table with a list of radio-buttons and above that is a drop down list of states. The drop down list of states is used to filter the table rows and when the selected item changes it posts-back to the server (via $("#frm").submit()). I don't want this to cause any validation to occur. Is there another function I can call besides submit(), or some other method?
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 need to validate two forms containing multiple input fields but want just one error message if any of the fields are left blank, the page is required to submit the users details (registration form). Also if any of these fields are left blank i don't want to be able to go to the next page on clicking the submit button
everything works up until the if(uscan == "n") so idk why it cant find what the value of the field is, because ive tested it and it seems to know even know what the value is.
whats the script for setting a Submit button to clickable in JS? Right now I've got a few submit buttons but i dont want the next one to be available until the last information was submitted.
.... PerformCheck fires when any submit button is click, which is logical but I'd like to prevent this if it was the cancel button that was clicked (and preserve form submission on pressing Enter whilst in the 'title' box.
For some reason when I have more that one button in the array only the first counts, meaning only when first is clicked is the button validation checked=true.[/quote]
