Re: mysql_query problem (more suggestions)

the problem is (as i understand it), the password function is a mySQL function and not a PHP function, therefor it works in the context of a mysql_query, but not as a PHP function (which your suggested code calls). Unless i have actually defined a function 'password()', calling 'echo password('$password') will (...does, i just double checked) causes a Fatal Error 'Undefined Function'. Nice try tho :) > From: Alvin Tan <alvintan@inscribe.com.my> > Date: Tue, 17 Jul 2001 13:07:45 +0800 > To: Brad Wright <bradwright@optushome.com.au> > Subject: Re: [PHP] mysql_query problem (more suggestions) > > Then, > > how about this: > > $pw = password($password); > $sql = "SELECT * FROM login_table where Pass ='$pw'"; > $testResult = mysql_query($sql) or die ("ouch"); > echo $sql; > > this way you can see what sql is talking to the db. > > @lvin > > At 12:59 PM 7/17/01, you wrote: >> password('$variable') when used in a mysql_query, uses the mysql function to >> convert the contents of $variable to an encrypted form for keeping in the >> database. >> >>> From: Alvin Tan <alvintan@inscribe.com.my> >>> Date: Tue, 17 Jul 2001 12:49:06 +0800 >>> To: Brad Wright <bradwright@optushome.com.au> >>> Subject: Re: [PHP] mysql_query problem >>> >>> Hi Brad, >>> >>> Shouldn't it be: >>> >>> $testResult = mysql_query("SELECT * FROM login_table where Pass = >>> '$password'") or die ("ouch"); >>> ^^^^^^^^^^^^^^ >>> >>> unless you have a function called password() you don't need to use the >>> $password variable within password('$password') >>> >>> HTH, >>> @lvin >>> >>> At 12:48 PM 7/17/01, you wrote: >>>> $testResult = mysql_query("SELECT * FROM login_table where Pass = >>>> password('$password')") or die ("ouch"); >>> >