I'd say $id is blank, not being passed in, or is equal to a nonexistant IDArt.
Maybe you should echo out your SQL and run it manually to see what's going on. J -- Jason Murray [EMAIL PROTECTED] Web Developer, Melbourne IT "Work now, freak later!" > -----Original Message----- > From: Dr. Shim [mailto:[EMAIL PROTECTED]] > Sent: Tuesday, April 02, 2002 2:41 PM > To: [EMAIL PROTECTED] > Subject: [PHP] Variable Appended To The End of a URL Is Not Working in > SQL Query > > > I have a variable which is appeneded to the end of a URL, like > > http://www.your_web_site.com/your_page/?your_variable=your_value > > This would return "your_value"; > > echo $your_variable; > > But this wouldn't work, and returns an error > > $sql = "SELECT * FROM fldField WHERE IDField = " . $id; > > > What could possibly be wrong? If you need more info, then > tell me. Here's my > code: > > <?php > $db = @odbc_connect('ReviewDatabase', 'root', '') or exit)("Error > occured:<br>$php_errormsg"); > $sql = "SELECT * FROM tblArt WHERE IDArt = $id"; > $cursor = @odbc_exec($db, $sql) or exit ("Error > occrued:<br>$php_errormsg"); > odbc_close($db); > ?> > > > > -- > PHP General Mailing List (http://www.php.net/) > To unsubscribe, visit: http://www.php.net/unsub.php > -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
