RE: [PHP-DB] mysql_query troubles

Brad, Have you defined the database variable you are connecting to? ie: $DB="your_database_name"; and then connecting: if(!($testResult = mysql_query($DB, "SELECT * FROM login_table where Pass = password('$password')"))) try the code snippet below and see if it spits an error at you - you will need to define the $DB variable. <? //error reporting function function DisplayErrMsg ( $message ) { sprintf("<BLOCKQUOTE><BLOCKQUOTE><BLOCKQUOTE><H3>%s</H3></BLOCKQUOTE></BLOCK QUOTE></BLOCKQUOTE>\n", $message); } //begin query if(!($testResult = mysql_query($DB, "SELECT * FROM login_table where Pass = password('$password')"))) //if we get an error print it { DisplayErrMsg(sprintf("internal error %d:%s\n", mysql_errno(), mysql_error())); exit(); } $num_rows = mysql_num_rows($testResult); print "$num_rows"; exit(); mysql_free_result($testResult); ?> -----Original Message----- From: Brad Wright [mailto:bradwright@optushome.com.au] Sent: 17 July 2001 06:31 To: php-db@lists.php.net Subject: [PHP-DB] mysql_query troubles Hi all', I have a mysql_query that reads: $testResult = mysql_query("SELECT * FROM login_table where Pass = password('$password')") or die ("ouch"); $num_rows = mysql_num_rows($testResult); the problem is that I keep getting a value of 0 for $num_rows when I know the table has 1 entry for PAss which equals password('$password'). The password I have used is just 'a' (before pasing throught the password function in the insert). I DO have a table named login_table I KNOW that there is an entry that matches 'password('$password)'.. ( i checked using phpMyAdmin) can anyone help?????? Thanks in advance, brad -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: php-db-unsubscribe@lists.php.net For additional commands, e-mail: php-db-help@lists.php.net To contact the list administrators, e-mail: php-list-admin@lists.php.net