MySQL Error Help!!

[Anon]

Hye sexy beautiful people i need help with this. Ok i have some code that connects to MySQL and here it is:

---

if ($cookie_user!="") {
$keys = array('userid', 'password', 'usergroupid', 'username', 'email');
foreach ($keys as $key) {
$result= mysql_query("SELECT * FROM user WHERE username='$cookie_user'");
$session[$key]= (mysql_result($result,0,$key));
}

}

so as you can see it works correctly when cookie_user has a correct name in it, what i need help with is getting it so that if its blank or the username is not in the db is will print a default like "guest" instead of 4 errors... can any one help me?

[mattw]

if ($cookie_user!="") {
$keys = array('userid', 'password', 'usergroupid', 'username', 'email');
foreach ($keys as $key) {
$result= mysql_query("SELECT * FROM user WHERE username='$cookie_user'");
if(mysql_num_rows($result==0) {
$session[$key]="guest";
} else {
$session[$key]= (mysql_result($result,0,$key));
}
}
}

Matt Wade
codewalkers.com

[Anon]

Thanks i tryed it and the information comes up, but i get this error 5 times

Warning: Supplied argument is not a valid MySQL result resource in /home/animeuto/public_html/rpg/global.php on line 11

and its in this line

    if(mysql_num_rows($result==0)) {

[mattw]

sorry...was typing that with my baby on my lap 🙂

if(mysql_num_rows($result==0) {

goes to

if(mysql_num_rows($result) == 0) {

Matt Wade
codewalkers.com

[Anon]

thanks it worked.