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2 Forms with Jquery


2 Forms with Jquery

By : Safal Trading
Date : November 24 2020, 03:01 PM
this will help First of all you call updateSingleOptionSelector_0_2() without a value and i think you throw an error there. This might be the reason other listeners doesn't work.
On the other hand change function works as expected.
code :
$('#first').change(function(){
	$('#second').val($(this).val())
});
$('#second').change(function(){
	$('#first').val($(this).val())
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<select id="first" class="select">
<option value="1">1 Option</option>
<option value="2">2 Option</option>
<option value="3">3 Option</option>
<option value="4">4 Option</option>
</select>

<select id="second" class="select">
<option value="1">1 Option</option>
<option value="2">2 Option</option>
<option value="3">3 Option</option>
<option value="4">4 Option</option>
</select>


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jQuery forms.js with multiple forms per page

jQuery forms.js with multiple forms per page


By : Allen
Date : March 29 2020, 07:55 AM
it fixes the issue I would like to submit information to a mySql database using php and ajax. , Change this code:
code :
jQuery('form').ajaxForm({ 
    target: '#noteReturn',
    success: function() { 
        $('#noteReturn').fadeIn('slow');
    } 
});
jQuery('form').ajaxForm({ 
    target: '#noteReturn',
    dataType: 'json',
    success: function(data) { 
        $('#noteReturn' + data.id).html(data.note).fadeIn('slow');
    } 
});
<?php
$note = $_POST['note'];
$id = $_POST['bid'];
$sql = "INSERT INTO notes (business_id, notes) VALUES ('$id', '$note')";
$result = mysql_query( $sql );
if($result) {
    echo " $note";
}
?>
<?php
$note = mysql_real_escape_string($_POST['note']);
$id = mysql_real_escape_string($_POST['bid']);
$sql = "INSERT INTO notes (business_id, notes) VALUES ('$id', '$note')";
$result = mysql_query( $sql );
if($result) {
    print json_encode(array("id" => $id, "note" => $note));
}
?>
Separate jquery listener for regular forms and forms containing a file input field

Separate jquery listener for regular forms and forms containing a file input field


By : mindaren
Date : March 29 2020, 07:55 AM
wish help you to fix your issue If your forms aren't dynamically injected into the page, then you don't need to use event delegation, and if your forms are dynamically injected into the page then that's not how you use event delegation. (You might want to read up on .on())
Assuming your forms are static in your page, then you can assign different forms depending if they contain input[type=file] using:
code :
var allForms = $('form'),
    uploadForms = allForms.has('input[type=file]');

uploadForms.on('submit', function(e) { ... });  // for forms with file uploads

allForms.not(uploadForms).on('submit', function(e) { ... });  // all other forms
// with uploads
$('body').on('submit', 'form:has(input[type=file])', function(e){ ... });

// all other forms
$('body').on('submit', 'form:not(:has(input[type=file]))', function(e){ ... });
Loading HTML forms through php then using JQuery to submit forms

Loading HTML forms through php then using JQuery to submit forms


By : Rob Davis
Date : March 29 2020, 07:55 AM
I hope this helps you . If you are writing out the forms in a for loop with php you can assign each submit button an id using the iterator, like submit_1, submit_2 etc and then you can have an on click handler in jquery using a selector contains, something like:
code :
$(document).on('click', 'input[id*="submit_"]', function() {
    //code goes here
    alert( $(this).prop('id') );
});
I have two forms in a page. How to set the jQuery script not to affect both forms buttons?

I have two forms in a page. How to set the jQuery script not to affect both forms buttons?


By : chair
Date : March 29 2020, 07:55 AM
will help you You have to use selectors like others say. Here's the modification to your jsfiddle: http://jsfiddle.net/baW53/1/
Basically, if its the initial call, we change the properties of all the buttons, i.e, buttons with class submit else we find the button which is a sibling of the dropdown and use its id.
code :
if (e) {
        var btn = "#" + jq(this).siblings('button').attr('id');
} else {
        var btn = ".submit";
}
jQuery: attach event to 2 forms only in page with 3 forms

jQuery: attach event to 2 forms only in page with 3 forms


By : Wael Aly
Date : March 29 2020, 07:55 AM
this will help I have a page with 3 forms, every form has it's unique id. I also have two scripts, one should be invoked with two forms but not the third and another only with the third. I can attach the script to the third only by selecting it like this: , You could use comma separator , to pass multiple selector :
code :
$('#first_form,#second_form')
$('#first_form,#second_form').on('submit', function(){
  alert('First or Second_form');
})
$('#third_form').on('submit', function(){
  alert('Third form');
})
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<form id='first_form'>
  <input type='submit' value='First form submit'/>
</form>
<br/>
<form id='second_form'>
  <input type='submit' value='Second form submit'/>
</form>
<br/>
<form id='third_form'>
  <input type='submit' value='Third form submit'/>
</form>
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