So what's the problem? The data doesn't appear in the table? Did you try to see if mysql_error() had anything in it after your query?
---John Holmes... > -----Original Message----- > From: Bill Arbuckle, Jr. [mailto:[EMAIL PROTECTED]] > Sent: Thursday, November 28, 2002 5:56 PM > To: [EMAIL PROTECTED] > Subject: [PHP-DB] 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