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 usi

Answer 1

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 2

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 3

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 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!