Form submit without refresh using jquery/ajax if page have more than one form

Question

Hi all I know that its very easy to submit a form without refreshing if there is only one form on the page but what about if there are more than one form on the page. Im using the following code for form submission and it works ok if there is only one form on the page. How can I change it to make it work when there are multiple forms on the page. Thanks in advance.

function processForm() { 
        $.ajax( {
            type: 'POST',
            url: form_process.php,
            data: 'user_name=' + encodeURIComponent(document.getElementById('user_name').value),
            success: function(data) {
                $('#message').html(data);
            }
        } );
}

<form action="" method="post" onsubmit="processForm();return false;">
<input type='text' name='user_name' id='user_name' value='' />
<input type='submit' name='submit' value='submit'/>
</form>
<div id='message'></div>

Answer 1

Just customize your function and add params like formid to get form data within the function to pass processForm("id of the form");

function processForm(formId) { 
    //your validation code
    $.ajax( {
            type: 'POST',
            url: form_process.php,
            data: $("#"+formId).serialize(), 
            success: function(data) {
                $('#message').html(data);
            }
        } );
    }

<form action="" id="form1" method="post" onsubmit="processForm('form1');return false;">
<input type='text' name='user_name' id='user_name' value='' />
<input type='submit' name='submit' value='submit'/>
</form>

<form action="" id="form2" method="post" onsubmit="processForm('form2');return false;">
<input type='text' name='user_name' id='user_name' value='' />
<input type='submit' name='submit' value='submit'/>
</form>
<form action="" id="form3" method="post" onsubmit="processForm('form3');return false;">
<input type='text' name='user_name' id='user_name' value='' />
<input type='submit' name='submit' value='submit'/>
</form>
<div id='message'></div>

Answer 2

What you have should work on multiple forms, however it would be better and a lot easier to debug if apply the event listener using jQuery instead:

$('form').submit(processForm); // listen to each form’s submit

function processForm(e) { 
    e.preventDefault(); // prevents default submit action
    $.ajax( {
        type: 'POST',
        url: form_process.php,
        data: 'user_name=' + encodeURIComponent(document.getElementById('user_name').value),
        success: function(data) {
            $('#message').html(data);
        }
    } );
}

HTML (without the ugly onsubmit attribute):

<form action="" method="post">
<input type='text' name='user_name' id='user_name' value='' />
<input type='submit' name='submit' value='submit'/>
</form>

Answer 3

Add form_id while calling processForm(form_id) and using the id serialize the form.

function processForm(form) { 
    $.ajax( {
        type: 'POST',
        url: form_process.php,
        data: $(form).serialize(),
        success: function(data) {
            $('#message').html(data);
        }
    } );
    return false;
}
<form action="" method="post" onsubmit="processForm(this)">
  <input type="text" name="user_name" id="user_name1" value="">
  <input type="submit" name="submit" value="submit" >
</form>
<form action="" method="post" onsubmit="processForm(this)">
  <input type="text" name="user_name" id="user_name2" value="">
  <input type="submit" name="submit" value="submit" >
</form>
<div id='message'></div>

jsFiddle

Answer 4

just thought I'd add to this, if you have multiple forms on one page, you can set a function to act on all forms independently by getting the form's action value. As follows (tested with jQuery 1.8.3):

 $('form').submit(function(){
   var action = $(this).attr("action")
   $.ajax({
     url: action,
     type:'POST',
     data: $(this).serialize(),
     success: function(){
            //alert message
        }                   
     })
   return false
 })

Hope this helps someone out!