I have a page with login form (on click, login form slides dowm) and underneath I have a link for registering a new user. With click on 'register new' link, new form pops-up and after validating each field, the submit event doesn't work - because there is a login form too and jQuery tries to trigger submit event of the first form.
How to trigger this specific onsubmit event - for the second form? I tried to hide a first form, but it didn't work and then I try to disabled it, which doesn't work as well.
(forms on separate pages works fine - validated with PHP and JS)
I think this is an easy thing to do .. but I cannot figure out how to do. Till now I overcome this problem, but I really like to find out how to make it work.
Thanks.
Unfortunatelly .. none of this answers works ...
Here is the code:
<div id="loginFormContainer">
<h2 id="loginShow" class="myriad_pro_bold_condensed_italic">Login</h2>
<form name="login_form" action="<?php if(isset($action) && !empty($action)) echo $action; ?>" method="POST" id="login_form" >
<fieldset>
<label>Your name </label>
<input type="text" name="username" maxlength="30" id="username_login" value="" class="required" placeholder="Enter Your Name" />
<label>Your password </label>
<input type="password" name="password" maxlength="16" id="password_login" value="" class="required " placeholder="Enter Your Password" />
</fieldset>
<input type="hidden" name="token" value="<?php if(isset($token)) echo $token; ?>" />
<input type="submit" name="submit-login" id="submit_login" value="Login" />
</form>
<div class="small">Register | Forgotten Password</div>
<div class="error"></div>
</div>
And JS ... should work like: if on register click: register windows pops-up, if on login the login form should slide down.
Right now, both of them slides ...I can remove the login form if I want to register, but the submit button from register form doesn't work.
$("#login_form").hide();
$("#loginShow").click(function(){
$('form#login_form .error').remove();
$("#login_form").slideDown("slow");
$('form#login_form').on("submit", function(event) {
event.preventDefault();
//code validation and ajax submit
}); //end form submit
});
$('#register').click(function(e) {
//$('#loginShow').remove(); //form is removed, but submit event still doesn't work
//if I completely remove login form from php page, then works fine
e.preventDefault();
$('form#register_form').click(function(event) {
event.preventDefault();
//
}); //end form submit
});//end register floating
Oh, yes, one final detail: on Chrome works fine :S, not in FF (v.10)
$(":submit").click(function(e){
$(this).closest("form").submit();
return false;
});
Try this:
var secondForm = $('form').eq(1);
secondForm.trigger('submit');
Just give the second submit button an id.
<input type='submit' id='submit2'>
Now for trigerring validation bind onclick event to the submit2 id
$('#submit2').click(function(){
//logic goes here
});
Related
I have Laravel 5.5 a page that has two forms but uses the same controller and method. First form is to cater for initial details but the second form is a search form. My search form works but only if you click the search button twice is there a way I could force to click once to submit that form.
View
<form name="FormSearch" id="FormSearch" method="post" action="{!! action(MyController#index',$customID) !!}">
<input type="text" name="searchDets" id="searchDets" value="" />
<input type="submit" class="btn btn-default" name="searchMe" id="searchMe" value= "Search Me"/>
</form>
Js
$('#FormSearch').click(function(e){
e.preventDefault();
$('#filterSearchForm').submit();
});
I would like my view page to submit once.
First of all, I couldn't find the element with id filterSearchForm, so, here is an edit that I made that submits the same form (i.e. FormSearch) when you click on it:
$('#FormSearch').click(function(e){
//e.preventDefault();
$('#FormSearch').trigger('submit');
});
I have used the trigger event that is fired which submits the form FormSearch when you click on the form FormSearch.
Here is it how it works:
$('#FormSearch').click(function(e) {
//e.preventDefault();
$('#FormSearch').trigger('submit');
});
$("#FormSearch").submit(function(e){
e.preventDefault();
alert('submitted');
});
#FormSearch{
background: black;
}
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<form name="FormSearch" id="FormSearch" method="post" action="{!! action(MyController#index',$customID) !!}">
<input type="text" name="searchDets" id="searchDets" value="" />
<input type="submit" class="btn btn-default" name="searchMe" id="searchMe" value="Search Me" />
</form>
<p>
** Click anywhere on the black area.
</p>
I hope this was helpful.
I have faced the same issue and it was due to js version . So please check your jquery.validate's version.
You can get more information here submit button twice click issue
I am Developing popup component for joomla site,
The Pop up Working Great , In my Popup i get phone number from user, i need to store that phone number to joomla database , but i am unable to call JFactory::getDBo(), when i call these method , popup was not working, i am in trouble , any help will be appreciate me.. thanxs in advance...
site/default.php
<script>
function openColorBox() {
$.colorbox({
innerWidth:500,
innerHeight:300,
iframe:true,
href: "subscribe.php",
overlayClose:true,
onLoad: function() {
$('#cboxClose').remove();
}
});
}
setTimeout(openColorBox, 1000);
</script>
site/subscribe.php
<body class="oneColFixCtr">
<div id="container">
<form name="Mail_list" action="#" method="post">
<p>
<label for="phone">Your Mobile Number </label>
<input type="tel" name="phone" id="phone" size="10" pattern="\d{10}" required />
<input type="hidden" name="date1" id="date1" value="<?php echo date('d.m.y'); ?>" />
</p>
<input type="submit" name="submit" value="Enter">
</form>
</div>
Your form is not posting the data anywhere when sumitted. Your action="#" will never allow the form to submit. Set your action to PHP_SELF if you need to submit it back to subscribe.php, then have a check in your subscribe.php that processes your form.
The better method would be to have your popup content in a hidden div and open that div instead of using an iframe. Use subscribe.php as your logic for saving the users data to the database. Using ajax to submit the form wouldn't be a bad idea either.
So what i want do is, have a form displayed on load where it asks user for their details such as name etc, then once the user clicks submit, i want that information to carry over to the next form, i know i have to hidden fields for that.
I want another form to be displayed after they click submit, this all has to be done using POSTBACK, so pretty much having 2 forms on one php page but only displaying the second one after the first has been submitted.
I know i can do this by creating two different php files and using header but i would like to learn how to do it via postback.
<form name="firstform" method="post" action="<?php echo $_SERVER['PHP_SELF']; ?> " onSubmit="return validator();">
<p>Please fill in the following form</p>
<p>Given Name* <input type="text" name="fname" id="fname"/><br/>
Middle Name <input type="text" name="mname"/><br />
Family Name* <input type="text" name="lname" id="lname"/><br />
Chosen Username* <input type="text" name="uname"/>
</p>
<p><input type="submit" value="submit" id="submit"/>
</form>
<form name="secondform" method="post" action="<?php echo $_SERVER['PHP_SELF']; ?> " >
Test* <input type = "text" name="test"/>
</form>
You can use jquery and Javascript if this is just a quick form
There is a nice form plugin that allows you to send an HTML form asynchroniously.
<script src="//cdnjs.cloudflare.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<script>
$(document).ready(function() {
// Hide all forms
$('form').hide();
// Show the first form
$('form').eq(0).show();
$('form').eq(0).on('submit', function() {
// Submit via ajax
// Unhide second form
$(this).hide();
$('form').eq(1).show();
return false;
});
});
</script>
I would validate that the first forms data is correct, then test to see if the form has been submitted without errors.
if (isset($_POST['submit']) && empty($errors)) {
// Output second form
} else {
// Output first form
}
You would either send the first set of data to the database or you can add them as hidden fields in the second form.
This is assuming you put any generated errors in to an array called $errors.
hi i am using a form for users subscription on the website with target is _new.
when i click on the submit button that the submitting message shows on the new window but the previous window still holding the data.
How to remove input fields data after submitting.
<form target="_new" action="samplepage.php" method="post">
<input type="text" name="inputtxt1" />
<input type="text" name="inputtxt2" />
<input type="submit" value="submit" />
</form>
Any suggestions???
Make the autocomplete - off
try this
<form target="_new" action="samplepage.php" method="post" autocomplete="off">
<input type="text" name="inputtxt1" />
<input type="text" name="inputtxt2" />
<input type="submit" value="submit" />
</form>
Reset form data using this
$('form').get(0).reset();
$(document).ready(function(){
$('form_name').submit(function(){
$('input[type="text"]').val('');
});
});
You can set value to null for text field using jquery on submit button click
$("input[type=submit]").click(function()
{
$("input[type=text]").val('');
});
EDIT:
I don't know is this a good approach, use blur instead of click event
$("input[type=submit]").blur(function()
{
$("input[type=text]").val('');
});
But it will meet your need.
I'm using a WordPress sidebar widget to capture email addresses. The thing is, it redirects after form submission. I want the visitor to stay on the page they were on after form submission with just a hidden div giving a successful signup message.
I've tried something with javascript like this --
<script type="text/javascript">
function showHide() {
var div = document.getElementById("hidden_div");
if (div.style.display == 'none') {
div.style.display = '';
}
else {
div.style.display = 'none';
}
}
</script>
And that works perfectly for showing the hidden div on submit, but the actual form then doesn't work :(
The form (with what I was trying to do) is like this --
<div id="wp_email_capture"><form name="wp_email_capture" method="post" onsubmit="showHide(); return false;" action="<?php echo $url; ?>">
<label class="wp-email-capture-name">Name:</label> <input name="wp-email-capture-name" type="text" class="wp-email-capture-name"><br/>
<label class="wp-email-capture-email">Email:</label> <input name="wp-email-capture-email" type="text" class="wp-email-capture-email"><br/>
<input type="hidden" name="wp_capture_action" value="1">
<input name="submit" type="submit" value="Submit" class="wp-email-capture-submit">
</form>
<div id="hidden_div" style="display:none"><p>Form successfully submitted.</p>
</div>
The problem is coming in somewhere between 'return false' and the form action (which is where the plugin's coder has made it redirect I think). If I remove 'return false', it redirects. With 'return false' there, the form doesn't work. I can't figure out a way to get the form to work but not redirect, ie. just show the hidden div, work, and that's it! No redirect :) Would appreciate your help.
I will show how to submit the form with jQuery, as this is what you have available to you:
First of all, you should make one small change to the form HTML. Namely, change showHide() to showHide(this), which will give showHide() access to the form element. The HTML should be:
<div id="wp_email_capture"><form name="wp_email_capture" method="post" onsubmit="showHide(this); return false;" action="<?php echo $url; ?>">
<label class="wp-email-capture-name">Name:</label> <input name="wp-email-capture-name" type="text" class="wp-email-capture-name"><br/>
<label class="wp-email-capture-email">Email:</label> <input name="wp-email-capture-email" type="text" class="wp-email-capture-email"><br/>
<input type="hidden" name="wp_capture_action" value="1">
<input name="submit" type="submit" value="Submit" class="wp-email-capture-submit">
</form>
<div id="hidden_div" style="display:none"><p>Form successfully submitted.</p>
</div>
The javascript to submit the form and display the div on successful submit is:
function showHide(form) {
var serial = $(form).serialize();
$.post(form.action, serial, function(){
$('#hidden_div').show();
});
};
What this does:
Serializes the form data, i.e. converts it to one long string such as wp-email-capture-name=&wp-email-capture-email=&wp_capture_action=1 that is stored in serial.
Submits the serialized data to the the form's action url (form.action)
If the form submit was successful, it runs the success handler, which is the third parameter to $.post(). This handler takes care of displaying the hidden div. I changed the code to use jQuery's .show() function, which takes care of browser inconsistencies.
Hope this is helpful.