Justin -
there's an error in your query. I think it should be:
$sql = "SELECT * FROM staff WHERE id='$username'";
Anyway, to see which is the problem just
after $result = mysql_query($sql);
HTH
Gianluca
JF> newbie to MySQL...
JF> I get this error:
JF> Supplied argument is not a valid MySQ
Hi all,
newbie to MySQL...
I get this error:
Supplied argument is not a valid MySQL result resource in /usr/local/
blah blah blah
>From this query:
$username = "juddy2";
$sql = "SELECT * FROM staff WHERE id=".$username;
$result = mysql_query($sql);
However, this works okay:
$username = "juddy
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