[PHP] Problem with Testing if the user has an account in my database

2002-08-15 Thread Sebastian Marcu

Hi All,

Please help me, I'm new to PHP and MySQL programming
and am really stuck on the attached code. 

I have a Flash front-end and am trying to query my
database for the 'registration' process to see if the
user already has an account. It's a very simple
registration string which compares the users email
address with the entries in the database. If the email
address is already present in the database an error
signal is sent to flash to notify the user. When the
email address is not present it gets added to the
database.

For some reason though it isn't working right and
keeps telling me that the 'Email' address is already
taken even though this entry is non-existant in the
database.

What am I doing wrong with my code below? Is there a
better way to check if the email address is already
contained in the database?

Any help would be much appreciated. Thank you all.

Regards,


Sebastian



___

require (config.inc.php);

$db = mysql_connect($dbserver,$nutzer,$passwort);

$myData = mysql_db_query($dbname,SELECT * FROM
learningZoneCustomer WHERE Email='$rEmail');

if ($myData  0)
{   
echo 'logInSig=1';

// error signal for flash to say that email is
already present in the database

}else{
// if the email address is not present in the
database add it to the database

mysql_db_query($dbname,INSERT INTO
learningZoneCustomer(Name,Surname,Email,Area,lChapter,VisitState,LearnState,Score,Date)
VALUES
('$rName','$rSurname','$rEmail','$rArea','$rlChapter','$rVisitState','$rLearnState','$rScore',
CURDATE()));

$pruf = mysql_affected_rows();

// Datensatz angelegt (Ja/Nein)

if ($pruf  0) {
echo 'regSig=1';
}


}

__


__

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[PHP] getting the IP address off a visitor

2002-07-08 Thread Sebastian Marcu

Hi there,

I am new to PHP and need some help.
I was wondering if there is a way to get hold of the IP address of a site
visitor with PHP. I'm trying to develop an interaction where the server
would recognise a new visitor from a returning visitor via the IP address.

Regards,


Sebastian





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