input not passing to php script

Question

I have a html form below:

    <section id="main" style="float: right;  text-align: left;  width: 83%;">   
        <article>
            <form>
                <p>Name:   <div><input id="name" type=text/></p></div><br \>
                <p>Phone:   <div><input id="phone" type=text/></p></div><br \>
                <p>Address:   <div><input id="address" type=text/></p></div><br \>
                <p>Username:   <div><input id="username" type=text/></p></div><br \>
                <p>Password:   <div><input id="password" type="password"/></p></div><br \>
                <p>Email:   <div><input id="email" type="text"/></p></div>
                <div><input id="submitDB" type="submit" value="Register"/></div>
            </form>
            <script type="text/javascript">
                      $('#submitDB').on('click', function() {
                          event.preventDefault();

                          $.ajax({
                               url: "registerDB.php",
                               data: { name: $('#name').val(), phone: $('#phone').val(), address: $('#address').val(), username: $('#username').val(), password: $('#password').val(), email: $('#email').val()},
                               type: "POST"
                          }).done(function(data) {
                              if(data === "true"){
                                  alert(data);
                              }
                              else
                              {
                                  alert(data);
                              }
                          });

                      });
           </script>
        </article>
    </section>

This form should submit the inputs and call a "registerDB.php" file which contains

    <?php
        require_once("config.php");
        require_once("MysqliDb.php");

        $DB = new MysqliDb($DBCREDS['HOSTNAME'], $DBCREDS['USERNAME'], $DBCREDS['PASSWORD'], $DBCREDS['DATABASE']);

        $name = $_POST["name"];
        $phone = $_POST['phone'];
        $address = $_POST['address'];
        $username = $_POST['username'];
        $password = $_POST['password'];
        $email = $_POST['email'];



        $data = Array ("name" => $name,
                       "phone" => $phone,
                       "address" => $address,
                       "email" => $email,
                    );

        var_dump($data);
       if($DB->insert('member', $data))
        {
                $user = Array ( 
                        "username" => $username,
                        "password" => $password,
                        );

                        var_dump($user);
                        die();
            if($DB->insert('member_login', $user))
            {
                echo "You are now registered!";
            }
        }   
        else
        {
            echo "Please Fill All Data And Resubmit";
        }





    ?>

However the username and password variables are not passing through. Is there something I am missing here? This does not have to be super "secure" I am working on a project for my System Analysis class and basically am working on making a website for a presentation.

biko · Accepted Answer · 2014-11-26 04:15:57Z

2

You need to add the name attribute to your inputs. Such as:

<form id="myForm">
    <p>Name:   <div><input id="name" name="name" type=text/></p></div><br \>
    <p>Phone:   <div><input id="phone" name="phone" type=text/></p></div><br \>
    <p>Address:   <div><input id="address" name="address" type=text/></p></div><br \>
    <p>Username:   <div><input id="username" name="username" type=text/></p></div><br \>
    <p>Password:   <div><input id="password" name="password" type="password"/></p></div><br \>
    <p>Email:   <div><input id="email" name="email" type="text"/></p></div>
    <div><input id="submitDB" type="submit" value="Register"/></div>
</form>

And use the following to submit the form #myForm with jquery. (make 'myForm' as the id of your form)

$("#myForm").ajaxForm({url: 'registerDB.php', type: 'post'})

TIP: use ajaxForm so as not to reinvent the wheel when creating ajax forms.

edited Nov 26, 2014 at 4:15

answered Nov 26, 2014 at 4:09

biko

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1 Comment

biko Over a year ago

Awesome, glad it did!

Justin Howard · Accepted Answer · 2014-11-26 04:10:10Z

2

You need to add the name attribute to your form inputs:

<form>
    <p>Name:   <div><input id="name" name="name" type=text/></p></div><br \>
    <p>Phone:   <div><input id="phone" name="phone" type=text/></p></div><br \>
    <p>Address:   <div><input id="address" name="address" type=text/></p></div><br \>
    <p>Username:   <div><input id="username" name="username" type=text/></p></div><br \>
    <p>Password:   <div><input id="password" name="password" type="password"/></p></div><br \>
    <p>Email:   <div><input id="email" name="email" type="text"/></p></div>
    <div><input id="submitDB" type="submit" value="Register"/></div>
</form>

answered Nov 26, 2014 at 4:10

Justin Howard

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Comments

Pupil · Accepted Answer · 2014-11-26 04:09:51Z

1

You are sending data with GET method in $.ajax:

type: "GET"

And in PHP, you are expecting:

$name = $_POST["name"];

Make it

type: "POST"

answered Nov 26, 2014 at 4:09

Pupil

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1 Comment

Paul Over a year ago

Yes, I noticed that after posting. That was an error on my part for not checking my code before uploading ( I was playing around with it to see if it was my POST that wasnt allowing it to pass). Thank you though!

Chris · Accepted Answer · 2014-11-26 04:20:39Z

0

I tested your HTML on my server and had no problems sending it to a basic PHP file. My PHP looked like this:

<?php
print_r($_POST);
?>

I suspect you're having an error with your PHP, perhaps with the connection to the SQL database. Try turning on error reporting to get some more information.

answered Nov 26, 2014 at 4:20

Chris

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