hi, i need help again

[Evan_Morgan]

hi i have no idea how to fix this, i always have problems with mysql arrays.

settype( $age, 'integer' );

$number = "".((($age+50)*10)-12)."";

$check = mysql_fetch_row("SELECT * FROM paid WHERE email = '$email'") or die(mysql_error());

print "$check[0]";

line 32 is the $check variable and i get this message when i try to access it

Warning: mysql_fetch_row(): supplied argument is not a valid MySQL result resource in /home/bigwager/public_html/domain/doemail.php on line 32

ive tried using fetch array as well

Any ideas?

Evan

[LordShryku]

You need to use mysql_query first....

[petitduc]

You just confused mysql_query and mysql_fetch_row.
the correct usage of these function would look like :

$result=mysql_query("SELECT * FROM paid WHERE email = '$email'") or die(mysql_error()); 
while($check=mysql_fetch_row($result)) {
   print $check[0]; 
}