[RESOLVED] PHP It just doesn't work!

[mkiller]

hi all,
i have a problem in my script.

<?php
if(isset($_POST['send']))
{
    mysql_connect("localhost","root","java");
    mysql_select_db('crimebay');

$naam = $_POST['naam'];
$pass = $_POST['pass'];
$pass1 = $_POST['pass1'];
$email = $_POST['email'];
$ip = $_SERVER['REMOTE_ADDR'];
$datum = date("j F Y");
 $db = "SELECT id FROM users WHERE id = $naam";
 $result = mysql_query($db);
if (empty($naam) || empty($pass) || empty($email) || empty($pass1))
{
    echo "Een of meer velden zijn niet ingevuld!";
}
elseif ($pass != $pass1){
echo "wachtwoorden zijn niet gelijk.";
}
elseif ($naam == admin || $naam == marcel || $naam == webmaster || $naam == crimebay){
echo "Je naam bevat illegale woorden.";
}
elseif(mysql_num_rows($result) == 1){
echo "Naam is al in gebruik";
}
else{
echo "Welkom bij CrimeBay " .$naam. ", nu kun je inloggen met je wachtwoord ( " .$pass. " ). Veel plezier op CrimeBay!";
$query= "INSERT INTO users VALUES('$naam','$pass','$email','$ip','0','$datum')";
mysql_query($query);
}
}
else
{
    echo "Fout, ga terug";
}
?>

now its about

 elseif(mysql_num_rows($result) == 1){
	echo "naam is al in gebruik";
	}

i can make more than 2 accounts with same name, but that is not what i want. it shoud say when i make a account with name that already exists Naam bestaat al. but i doesn't work.

i hope somebody can help me out in this script, that would be greatfull.

Greets, Marcel.
btw. Sorry for my bad english im dutch.

[bpat1434]

First thing I notice is this line:

elseif ($naam == admin || $naam == marcel || $naam == webmaster || $naam == crimebay){ 

Those strings (admin, marcel, webmaster, crimebay) should be in quotes. E.g.

elseif ($naam == 'admin' || $naam == 'marcel' || $naam == 'webmaster' || $naam == 'crimebay'){ 

You could probably make this a little easier to maintain if you turned those into names into an array of "reserved" names:

$reserved = array('admin', 'marcel', 'webmaster', 'crimebay');
/**
 * The original if statements here....
 */
elseif(in_array($naam, $reserved)) {
    /* The name is reserved */
}

I would not limit yourself to this 1 row logic:

elseif(mysql_num_rows($result) == 1){

For testing purposes, try out:

elseif(mysql_num_rows($result) > 0){

[Bjom]

change your DB table. Add a unique key to the name - that will prevent doublettes. Then you can check if the insert succeeds - if not, you know someone used an already existing name...and you can react with an appropriate message.

[mkiller]

1st - thanks for fast reply!
2nd - it doesn't work..
i tryd it and i stil can make 2 acc with same name. maybe its because i have php 5.2.10 .
because

	
elseif (in_array($verbod)){
	echo "Je naam bevat illegale woorden.";
	}

doesn't work now anymore too
maybe i should like "downgrade" php or such thing.
anyway if you guys have any other ideas i would like to hear them 😃

[Bjom]

"it doesnt work"

well great.

Make it work. All I can say.

For a more specific answer, be more specific in your questions.

Check your DB again. If a field is set to be indexed with a unique key it does not allow doublettes under any circumstances...

Bjom

[Installer]

mkiller;10923184 wrote:

	
elseif (in_array($verbod)){
	echo "Je naam bevat illegale woorden.";
	}

doesn't work now anymore too

I'm surprised it ever worked, since in_array() requires two arguments.

For your query, try this:

$db = "SELECT id FROM users WHERE id = '$naam'";

[mkiller]

jee sorry, but i get really frustrated cause i wont work.

but now it works!
in_array() needs 2 aruments and i had only 1
and $naam in $db = "SELECT id FROM users WHERE id = $naam";
is now in ' '
thanks al for youre fast reply
now, keep on the good work!