PHP/MySQL - Update Database/Run PHP Script on Button Click

Question

I have a 'registration' page in PHP and I would like the script to run when an HTML button is clicked.

The PHP basically checks if all fields are filled, checks if the password and email confirmations are the same and saves to the database.

This is the code:

<?php
$Name = isset($_POST['Name']);
$Surname = isset($_POST['Surname']);

$Username = isset($_POST['Username']);

$Email = isset($_POST['Email']);
$C_Email = isset($_POST['C_Email']);

$Password = isset($_POST['password']);
$C_Password = isset($_POST['c_password']);

$SecQ = isset($_POST['SecQ']);
$SecA = isset($_POST['SecA']);


$con = mysql_connect('localhost', 'admin', 'storefile1234');
mysql_select_db ("storefile");

$check_username = mysql_query("SELECT FROM users WHERE username = '$Username'");
$check_email = mysql_query("SELECT FROM users WHERE Email = '$Email'");


if (!$con)
        {
        die('Could not connect: ' . mysql_error());
        }

if ($Name == null || $Surname== null || $Username == null || $Password == null || $C_Password == null || $Email == null || $C_Email == null || $SecQ == null || $SecA == null ) {

    echo "Missing details. Please enter all fields.";


} else {

    if(mysql_num_rows($check_username) != 0 && mysql_num_rows($check_email) != 0)
            {
            echo "Username/Email already exists";
            }
            if  ($Email == $C_Email && $Password == $C_Password) {

                $query = "INSERT INTO users (Username, Name,Surname, Password, Email, SecQ, SecA) VALUES ('NULL', ".$Username."', ".$Name."', ".$Surname."', ".$Password."', ".$SecQ."', ".$SecA."', ".$Email.')"';

                mysql_query($query) or die ('Error registering.');

                echo "Greetings, ".$Name.", you have been registered. ";

    }  else {

        echo "Error registering your account. Please try again.";
            }

 }


?>

Also, is it recommended?

Whenever I run this page Missing details. Please enter all fields. displays, without having entered any details.

How do you do this?

offtopic: dont use mysql_* functions. Beter look on PDO with prepared statesments. — Stranger
– Stranger, Commented Dec 5, 2012 at 15:55

Stranger · Accepted Answer · 2012-12-05 16:05:37Z

2

You tying to get values by isset($_POST['Username']); and like this functions...
But documentation says: Returns TRUE if var exists and has value other than NULL, FALSE otherwise.

So check on true, nut null. And escape your POST data after.

You can do like this:

$Name = isset($_POST['Name']) ? mysql_real_escape_string($_POST['Name']) : null;

P.S. Please again. Do not use mysql_* function. They are DEPRECATED.
Look on PDO (or mysqli_*)

edited Dec 5, 2012 at 16:05

answered Dec 5, 2012 at 15:58

Stranger

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3 Comments

Benjamin Paap Over a year ago

The check would not solve his problem because he is reusing the variables he assigns the return of isset to in his sql statement.

Brian Over a year ago

@Stranger So basically I need to fix the isset() functions and it will work on button click?

Stranger Over a year ago

@Brian Yes. You can just check if they exist or not (true|false), but you dont get value of them. Benjamin answer isnt good enough, but solevs problem.

DanielX2010 · Accepted Answer · 2012-12-05 16:08:48Z

0

For the issue of printing that message when you first load the page, use the array_key_exists function to test if the user has already submited something before checking if any field is null. Something like this:

if (array_key_exists('Name', $_POST) || array_key_exists('Surname', $_POST) || ... ) 
    if ($Name == null || $Surname== null || ... ) 
        echo "Missing details. Please enter all fields.";

Observation: you cannot use the isset function for the same purpose since, according to php documentation, it "determine if a variable is set and is not NULL"

answered Dec 5, 2012 at 16:08

DanielX2010

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1 Comment

Brian Over a year ago

that seemed to work, however, XAMPP is somehow forbidding me from accessing the DB, givinig me this error

Access forbidden! You don't have permission to access the requested object. It is either read-protected or not readable by the server. Error 403

The host, username and pass are correct for thr DB.

Benjamin Paap · Accepted Answer · 2012-12-05 15:58:28Z

0

You misuse isset

Try something like this:

$Name = null;
if (isset($_POST['Name'])) {
    $Name = $_POST['Name'];
}

isset is only to check if a value is set.

answered Dec 5, 2012 at 15:58

Benjamin Paap

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2 Comments

Brian Over a year ago

So you need to use that function for each variable ?

Stranger Over a year ago

And dont forget to ESCAPE post data!

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