Hi all, Im getting mighty frustrated with this peice of code. It checks the
first condition no probs, but the second IF statement doesnt work. So it
will stop if it finds a du[plicate login, but cannot find duplicate
passwords. the $num_rows2 variable always gets the value '0' even when there
is an existing password the same as the new user-entered value.
I hope its a really simple syntax problem that i cant see. The slight
syntaxic (is that a word?) differences between the 2 IF statements are due
to 'trying anything' to get it working....
I hope someone out there can help.
Thanks in advance,
Brad
// ************* Check for existing duplicate login
$result = mysql_query("SELECT * FROM Login_TB where Login='$Login'",$db) or
die ("ouch");
$num_rows = mysql_num_rows($result);
// echo $num_rows;
if ($num_rows != 0)
{
session_register("errorMsg");
$errorMsg = "Username taken.";
echo $errorMsg;
exit;
}
// ************* Check for existing duplicate password
$query2 = "select * FROM Team_Login_TB where Pass = password('$Pass')";
$result2 = mysql_query($query2,$db) or die ("ouch2");
$num_rows2 = mysql_num_rows($result2);
echo $num_rows2;
if ($num_rows2 != 0)
{
session_register("errorMsg");
$errorMsg = "Password taken.";
echo $errorMsg;
exit;
}
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