Question

I am trying to use jquery’s form plugin from http://www.malsup.com/jquery/form/#ajaxSubmit and .ajaxsubmit to submit my data in a form however I am not really sure what .ajaxsubmit is passing and how I can read this in my php file.

I have a validate function

function validate(formData, jqForm, options) { 
    alert('About to submit: \n\n' + queryString); 
    return true; 
}

that shows queryString which is

first=testfirstname&last=testlastname&age=90

when I use .ajaxsubmit, nothing happens as listed in my script below.

$(document).ready(function() {          
    var options = { 
        target:        '#output1', 
        beforeSubmit:  validate,
        success:       showResponse
    };

    //submission 
    $('#myForm').submit(function() { 
        $(this).ajaxSubmit(options);      
        return false; 
    });
});

My form is

<form action="comment.php" method="post" id="myForm">

I was wondering what format is the data being sent, would I do something with

$_REQUEST['first'];

and also how would I also pass in an addition value from the $_SESSION?

Thanks

Answer 1

As far as I know, the jQuery plugin actually sends the plugin data as POST-data to PHP (similar to setting method="post" on your <form> tag). You can access it like this:

$_POST['name_of_field_in_form'];

The name_of_field_in_form is just the name of a field, for example if you have this code <input name="email" type="text" />, you could access it via $_POST['email'];.

About your second query, not sure what you mean, but you can use session_start(); to create a session and after that $_SESSION acts like a ‘normal’ array.