Re: Whats wrong with this code?

On Tue, 17 Jul 2001 12:37, Brad Wright wrote: > > From: Rasmus Lerdorf <rasmus@php.net> > > Date: Mon, 16 Jul 2001 19:56:29 -0700 (PDT) > > To: Brad Wright <bradwright@optushome.com.au> > > Cc: PHP General List <php-general@lists.php.net> > > Subject: Re: [PHP] Whats wrong with this code? > > > >> 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. > >> > >> $result = mysql_query("SELECT * FROM Login_TB where > >> Login='$Login'",$db) or die ("ouch"); > >> > >> $num_rows = mysql_num_rows($result); > >> > >> $query2 = "select * FROM Team_Login_TB where Pass = > >> password('$Pass')"; $result2 = mysql_query($query2,$db) or die > >> ("ouch2"); > >> $num_rows2 = mysql_num_rows($result2); > > > > Your second query must not be returning any rows. Print out $query2 > > and paste the string you see into the command line mysql tool and > > execute the query. I bet it will return 0 rows. > > > > -Rasmus > Rasmus, > you are dead right, it is returning 0 rows. I checked that before > submiytting the original post. Sorry, i should have mentioned it. I > guess the question is, why is it returning no rows? This might be too obvious, but are you looking in the correct table? Your second query searches a different table from the first? Alternatively, you might be getting something you don't expect in $Pass. -- David Robley Techno-JoaT, Web Maintainer, Mail List Admin, etc CENTRE FOR INJURY STUDIES Flinders University, SOUTH AUSTRALIA 1200 bps used to seem so fast