Hi you need quotes around the values mysql_query("insert into tblquotehits (qoption,referer,browser,remoteaddr,dt) values ('$quotetype','$referer','$browser','$remoteaddress','$currentdtstring')",$d b); HTH Peter ----------------------------------------------- Excellence in internet and open source software ----------------------------------------------- Sunmaia Birmingham UK www.sunmaia.net tel. 0121-242-1473 International +44-121-242-1473 ----------------------------------------------- -----Original Message----- From: Bill Arbuckle, Jr. [mailto:bill@wingshooter.net] Sent: 28 November 2002 22:56 To: php-db@lists.php.net Subject: mysql problem Hello all! I am new to php and I am trying to update a mySQL table using php v4.2.3 for Windows. I am using the command: mysql_query("insert into tblquotehits (qoption,referer,browser,remoteaddr,dt) values ($quotetype,$referer,$browser,$remoteaddress,$currentdtstring)",$db); I have read the documentation and found some notes on the topic but none seemed to solve my problem. I have tried other variations of the command such as inserting a semicolon at the end of the insert statement before the closing double quotes. All variables are defined and all columns are actually columns in the table ... I have checked all of this. I have also returned the result to a variable and echoed it to the screen ... the result printed is "Resource id #2". This is driving me crazy as I know it must be something quite simple that I am overlooking. Any help is greatly appreciated! TIA Bill -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php