Re: [PHP] Variable Appended To The End of a URL Is Not Working in SQL Query

2002-04-02 Thread Dr. Shim
sults? -Original Message- From: Dr. Shim [mailto:[EMAIL PROTECTED]] Sent: Monday, April 01, 2002 10:48 PM To: [EMAIL PROTECTED] Subject: Re: [PHP] Variable Appended To The End of a URL Is Not Working in SQL Query Hmm, run it manually? I'm a newbie, so, could you explain how I'd do th

RE: [PHP] Variable Appended To The End of a URL Is Not Working in SQL Query

2002-04-02 Thread Rick Emery
:48 PM To: [EMAIL PROTECTED] Subject: Re: [PHP] Variable Appended To The End of a URL Is Not Working in SQL Query Hmm, run it manually? I'm a newbie, so, could you explain how I'd do that? =) "Jason Murray" <[EMAIL PROTECTED]> wrote in message [EMAIL PROTECTED]">

Re: [PHP] Variable Appended To The End of a URL Is Not Working in SQL Query

2002-04-01 Thread Dr. Shim
The "your_variable" thing was just an example. It actually uses $id (an a real URL). "Tyler Longren" <[EMAIL PROTECTED]> wrote in message 003001c1da02$f46e65b0$0101a8c0@nightengale">news:003001c1da02$f46e65b0$0101a8c0@nightengale... Well, there's not $id variable. And you're trying to select it.

Re: [PHP] Variable Appended To The End of a URL Is Not Working in SQL Query

2002-04-01 Thread Tyler Longren
Well, there's not $id variable. And you're trying to select it. Should be this: $php_errormsg"); // next line has changed $sql = "SELECT * FROM tblArt WHERE IDArt = $your_variable"; $cursor = @odbc_exec($db, $sql) or exit ("Error occrued:$php_errormsg"); odbc_close($db); ?> Tyler Longren Cap

Re: [PHP] Variable Appended To The End of a URL Is Not Working in SQL Query

2002-04-01 Thread Dr. Shim
Hmm, run it manually? I'm a newbie, so, could you explain how I'd do that? =) "Jason Murray" <[EMAIL PROTECTED]> wrote in message [EMAIL PROTECTED]">news:[EMAIL PROTECTED]... 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

RE: [PHP] Variable Appended To The End of a URL Is Not Working in SQL Query

2002-04-01 Thread Jason Murray
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