mysql problem
| From: | Bill Arbuckle, Jr. | Date: | Thu, 28 Nov 2002 22:56:17 +0000 |
| Subject: | mysql problem | ||
| Groups: | php.db | ||
| Request: | Send a blank email to php-db+get-23723@lists.php.net to get a copy of this message | ||
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