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
> Subject: [PHP-DB] mysql problem
> Hello all!  I am new to php and I am trying to update a mySQL table
> 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
> seemed to solve my problem.  I have tried other variations of the
> such as inserting a semicolon at the end of the insert statement
> 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
> 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
> I
> am overlooking.  Any help is greatly appreciated!  TIA
> Bill
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