RE: [PHP] Help?

Hi everyone, Thanks for previous emails everyone, I have tried Julie's solution but I still didn't work. I have tried $result = mysql_query("Select * from $Game", $db) $result = mysql_query("Select * from '$Game'", $db) They all didn't work and gave me the same error. What is wrong with my code, please help? Thanks. Calvin -----Original Message----- From: Julie Meloni [mailto:julie@thickbook.com] Sent: Monday, July 31, 2000 12:09 PM To: Wee Chua Cc: PHP (E-mail) Subject: Re: [PHP] Help? > <option value"Emp">Employee</option> > <option value"Comp">Company</option> You're going to want the "=" in there... <option value="Emp">Employee</option> <option value="Comp">Company</option> So, now $Game will = either "Emp" or "Comp". Assuming you've made a connection and selected a database, and $db is the name of the link identifier you created with that selection, your SQL statement of "select * from $Game" would be just fine, provided that variable tracking is on and $Game becomes "Emp" or "Comp". > $result = mysql_query("Select * from $Game",$db); You can add a die() to this, to print a nice error: $result = mysql_query("Select * from $Game",$db) or die(mysql_error()); - julie +------------------------------------------------+ | Julie Meloni (jcm@i2ii.com) | | Tech. Director, i2i Interactive (www.i2ii.com) | | | | "PHP Essentials" & "PHP Fast & Easy" | | http://www.thickbook.com/ | +------------------------------------------------+