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
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-----Original Message-----
From: Bill Arbuckle, Jr. [mailto:[EMAIL PROTECTED]]
Sent: 28 November 2002 22:56
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
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