PHP form processing using JQuery not Working

Question

I'm processing a form using PHP and JQuery and it's not processing the way it should normally process. My HTML is:

    <form action="" method="POST" name="forgot_pass_form">
      <label><h3>Please Enter your email!</h3></label><br>
      <input type="text" name="email_pass" id="email_pass" />
      <input type="submit" name="submit" id="submit" value="Submit" />
    </form>

My JQuery code which is in the same page as HTML code:

<script type="text/javascript">
    $(document).ready(function($) {

        var email = $('#email_pass').val();

        $("#submit").live('click', function(){
                $.post('email_password.php', 
                {
                'email_pass' : email
                }, 
                function(data) {
                    alert(data);                
                });

                return false;
            });
    }); 
</script>

My PHP code which is on email_password.php:

<?php


    $email = $_POST['email_pass'];

    if(empty($email)){
        echo  'empty';  
    } else {
        echo 'Not'; 
    }
?>

The problem here is it always alert me with empty even if I've entered something in the text box

I even tried this way:

 if(empty($_POST)){
    echo 'empty';    
   } else {
    // process the form 
    $email = $_POST['email'];
    if(empty($email)){
        echo 'Email is empty';
    } else {
        echo 'Email is not empty';  
    }
}

By trying this way, it alerts me with Email is empty. Please help

what your network console says , is it posting "post" params? — Dimag Kharab
– Dimag Kharab, Commented Mar 27, 2014 at 12:23
I think it's 1.9 or 1.10. The problem was that I was not using var email inside the .live() method. It's solved :) — user3293145
– user3293145, Commented Mar 27, 2014 at 12:27
@user3293145 Please check out my answer here for a thorough AJAX example — MonkeyZeus
– MonkeyZeus, Commented Mar 27, 2014 at 12:30
@MonkeyZeus What to use for .attr('action') if the form is submitting to itself using form action=""? — user3293145
– user3293145, Commented Mar 27, 2014 at 12:36

wrivas · Accepted Answer · 2014-03-27 12:19:03Z

3

When you try to capture a field value, it must be the same name. The name you have in your form is 'email_pass' and in your PHP code you are waiting for 'email'

This should be:

if(empty($_POST)){
    echo 'empty';    
   } else {
    // process the form 
    $email = $_POST['email_pass']; // The field form name
    if(empty($email)){
        echo 'Email is empty';
    } else {
        echo 'Email is not empty';  
    }
}

Sorry for my english :s

answered Mar 27, 2014 at 12:19

wrivas

5096 silver badges12 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

user3293145 Over a year ago

Sorry that was a silly mistake, but I still get the same response as before, nothing changes

Ananth · Accepted Answer · 2014-03-27 12:17:31Z

2

use this code

<script type="text/javascript">
    $(document).ready(function($) {



        $("#submit").live('click', function(){
 var email = $('#email_pass').val();
                $.post('email_password.php', 
                {
                'email_pass' : email
                }, 
                function(data) {
                    alert(data);                
                });

                return false;
            });
    }); 
</script>

answered Mar 27, 2014 at 12:17

Ananth

1,5823 gold badges17 silver badges36 bronze badges

5 Comments

Ananth Over a year ago

add the var email = $('#email_pass').val(); inside the $("#submit").live('click', function(){

AyB Over a year ago

What difference does it make? @user3293145 although not the solution, .live() is deprecated, your should upgrade your jQuery and use .on().

Ananth Over a year ago

@ I Can Has Cheezburger. i know but he is using old version

user3293145 Over a year ago

I'm using CMS and even after upgrading JQuery, .on() doesn't work most of the time while .live() does

user3293145 Over a year ago

Yeah, it's asking me to wait for 2 minutes @Ananth

Sam Miller · Accepted Answer · 2014-03-27 12:21:13Z

2

The issue is that your making a $_POST call with jQuery and in that function you "post" under the new name email_pass. Update your PHP file to reference $_POST['email_pass'] and you'll be good to go.

answered Mar 27, 2014 at 12:21

Sam Miller

1665 bronze badges

Comments

Sridhar R · Accepted Answer · 2014-03-27 12:24:36Z

1

Use 0n() and get email value inside the click function

<script type="text/javascript">
    $(document).ready(function($) {

          $("#submit").on('click', function(){
                var email = $('#email_pass').val();
                $.post('email_password.php', 
                {
                'email_pass' : email
                }, 
                function(data) {
                    alert(data);                
                });

                return false;
            });
    }); 
</script>

answered Mar 27, 2014 at 12:24

Sridhar R

20.4k7 gold badges41 silver badges35 bronze badges

3 Comments

MonkeyZeus Over a year ago

Listening for the click of the button is improper. You must listen for form submission to properly catch the event in ALL browsers. $('form[name="forgot_pass_form"]').on('submit', functione(){});

user3293145 Over a year ago

Would it make difference on different browser if .on('click') is used instead of .on('submit')? @MonkeyZeus

MonkeyZeus Over a year ago

@user3293145 that is exactly what I said, yes. Listening for the click of a button is a different event compared to the submission of a form. Even though click is working for you, it is not 100% reliable. Listening for a submit is 100% reliable.

Collectives™ on Stack Overflow

PHP form processing using JQuery not Working

4 Answers 4

1 Comment

5 Comments

Comments

3 Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

4 Answers 4

1 Comment

5 Comments

Comments

3 Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related