Is there a way to validate a form using a link and NOT a submit button? I have a link on my form that I'm using as a submit button to do some ajax posting. However, I need to validate some of the input fields before the data is sent to the database.
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I am having the following issue, when I try using an image to post Icannot get the validation to run. When I click the button the formsubmits and nothing is checked. I've been searching all over the web One way of coding doesnt work and the other
does. I have included both below.
$().ready(function() {
var container = $('div.container');
var validator = $("#CustForm").validate({
[code]....
Here is my handler:
Here is my button:
How do i trigger a validate when the submit button is clicked?
All of my validations work while I am working with the form but the button does zilch.
I'm using mailchimps signup box (they don't have a decent forum to ask on) on my website and want to adjust the submit button and change it to a normal link. Here's the button that submit's the form:
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I currently use javascript in an anchor tag to submit a form using
<a href="#" onClick="document.myformname.submit(); return false;">Submit</a>
The only problem is, JQuery doesn't acknowledge that my form has been submitted.
I've got a pretty basic form that i need to be able to "submit" by simply loading it in the browser address, not click the submit button. is there a way to do this? Here is the code:
<form name="newform" action="/dir/submition.cfm?z=1&Myid=1" method="post">
<input type="hidden" name="userID" value="27566">
<input type="submit" value="Submit">
</form>
if i load just http://www.mysite.com/dir/submition.cfm?z=1&Myid=1, it doesnt do anything because it doesnt have the userID variable, but im not sure how to include that into the address...
I have a requirement that when a particular link is clicked, it submits back to the server. This is the same as clicking on a submit button. But the submit button is a link.
View 2 Replies View RelatedI would like my form submit button to show as a link instead of a button, where it will be a link and pass the city value to the action page.
View 3 Replies View RelatedI have recently needed to replace the submit buttons on my forms with clickable links instead. I wanted the replacement to be completely compatible with my existing PHP server-side code. This means that where I checked for a button having been pressed on the server side, the same check would continue to work, and the same form elements would have been submitted.Stripping away issues of screen formatting, I changed my button code generator to produce this
Code HTML4Strict:
<a href="javascript:submitMe('buttonName');">buttonLabel</a>
instead of this
Code HTML4Strict:
<input type="submit" name="buttonName" value="buttonLabel">
and in javascript I wrote this function
Code JavaScript:
function submitMe(val)
{
// create a form element that will show up in $_POST with the same
[code]....
I am using the Jquery validationss plugin for my form validations. NowI have a situation where I have two submit buttons in a single form(say button A and button B). For button A, I want to run validationsbut for button B I don't want to run the validations and submit theform without validating.. How can i do this?
View 2 Replies View RelatedI have 3 radio buttons with the same name ("myradio"). The value of each is a different URL (eg. value="http:[url]...)I have a link underneath these (<a href="#">Send</a>). I would like the selected radio button's value to be inserted into the href of the link, and if the user selects a different radio button that this changes.
View 1 Replies View Relatedim trying to learn how jquery ajax function works by validating some form after submit is clicked.
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the alert didnt pop until i added the registration_ok = false; right before, so im assuming it has something to do with the ajax functions's scope.... why does the ajax function not affect the registration_ok variable outside the function? and how should i solve this problem? any better way of doing it?
I am a newbie to jQuery and still learning the ropes. I would greatly appreciate your help in resolving the following issue: I am using jquery.validate.js plugin to validate a form.
The form has two groups of radio buttons with names ( g1 and g2 ). It also has other elements like textboxes and select boxes.
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In the following .submit function, I am attempting to grab the value of the selected option in the facilityCodes dropdown list when I click the submit button and then during the submit function,select the facilityCode again in the dropdown list when the page reloads and then disable the list so that the user cannot change it.However,the situation is when I reload the page after the submit button is clicked the dropdown defaults to the first option and the list is enabled.I apparently am not understanding how.submit works so that I'm selecting the option I'm defining in my code below and then disabling the list AFTER the page reloads due to an error on the page. My question is how can I accomplish this?Here is my code:
$(function() {
$("#ARTransferForm").submit(function() {
var msgsCount = 0;[code]....
I have a page with many forms that I need to change from a post to an ajax call. That part is working, no problem, but now I want to disable the submit button while it's waiting on the server response and then re-enable it when the response comes back.
Here's what I have:
$(function() {
$('form').each(function() {
$(this).submit(function(){
[code]...
I can't figure out what my selector should be to get the submit button of the form that's being submitted. What should I be using instead? Also, if the call errors out, I'd like to just post the form as usual.
I am using a validating form plug in for jquery and I have a question about it. Let this function will be an ex.:
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'e' is the name attribute of one form element, but can I choose more elements using jquery (CSS) rules like this: input[name*=e] or how can I do something similar?
I have this as my Dialog Javascript
$( "#dialog-form<?php echo $diagnumber; ?>" ).dialog({
autoOpen: false,
height: 300,
[code]....
I have a form that has multiple radio button groups and i want it to validate so that if any button is selected in one group you cannot make a selection in the other groups.
View 1 Replies View RelatedI have a group 2 Radio buttons, if the user selects the 2nd Radio button then they need to enter some data into a text field... The following only works for either Radio buttons... How can I specify it to be only the 2nd Radio button?
if(!document.form.radio1.checked && document.form.q3.value=="")
{
theMessage = theMessage + "
--> You selected no we need a reason)";
}
The application I'm helping develop is about 10 pages long and allows users to move back and forth, updating previous selections. Certain pages use Javascript to check current selections and show/hide divs based on those selections. Most of these fields are radio buttons and use simple onClick calls. However, one of these requires that 2 fields be checked for a specific value and if they are both true, display a div.
One problem though... for other reasons these form field names all begin with either a "r^" or a "t^" signifying it as either a text field or radio button. I can get the check to work fine without the "r^" but with it I'm lost. The carrot seems to be causing problems for me and I've tried many different combinations of things to get it to work. Code:
I have a form that makes visible a text field when a specific radio button is checked. That works. Then I validate to see if a radio button is checked before I let the form be submitted. That works.
BUT - I cannot seem to get it to validate the form field that is made visible to see if it has content or at least the proper content when the form is submitted. Can you tell me what I am doing wrong please?
I can give more info on what kind of validation I want done if needed, but I think my code speaks for itself Code:
how can i stop validating a value if i press the cancel button in confirm().. because in my code if i press the cancel button it still passes a value
View 5 Replies View RelatedI copied codes on internet and arranged them according to my knowledge of JQuery (which is very shallow).. The code works fine sometimes.... but not all the time. Can anyone help me to find out why the dialog box does not appear everytime I put wrong username and password?
Here is the JQuerycode:
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Because I am using ui-darkness as the page theme, so at form tag, I don't have defaultbutton="but_submit" tag. If I add it back, I will get error while compiling..
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.
View 2 Replies View RelatedThe following will submit the form data to popup by clicking the submit button. I want it will submit the form automatically to the popup, there is no submit button in this page. Basically this page should not show up.
<script type="text/javascript">
function submitmyform(f) {
f.target = 'foo'
window.open('',f.target,'menubar=no,scrollbars=no, width=800,height=800');
f.submit();
return false;
}
</script>
<form name="myform" action="popup.asp" target="_blank" method="post"
onsubmit="return submitmyform(this);">
<input type="hidden" name="item" value="item"/>
<input type="submit" value="submit to popup"/>
</form>
I have a form without a submit button. It gets submitted programmatically with document.form.submit().
What I need is to be able to disable this form's submit capability on page load and then reenable it at some later point. Remember there is no 'input' button element.
Code:
What I've tried so far is like this:
savedSubmit = document.inputs.submit;
document.inputs.submit = None;
then later:
document.inputs.submit = savedSubmit;
but this does not work. How can I do this?
