Ahhh! quote that ID number before using it in a query! :) // for mysql mysql_quote($_GET['ID']);
--- Scott Hurring Systems Programmer EAC Corporation [EMAIL PROTECTED] Voice: 201-462-2149 Fax: 201-288-1515 > -----Original Message----- > From: 1LT John W. Holmes [mailto:[EMAIL PROTECTED]] > Sent: Wednesday, June 05, 2002 4:29 PM > To: Igor Portnoy; [EMAIL PROTECTED] > Subject: Re: [PHP] Passing variable to new page and pulling > the rest of > info from database > > > SELECT * FROM table WHERE ID = $_GET['ID'] > > Then create a page to display all of that information. Look > at the mysql > functions and learn some PHP. We can't help you without > knowing what's in > your table and how you want it displayed, etc... > > So keep learning and reading and you'll figure out how to do it. Your > question is way to broad for any help... > > ---John Holmes... > > ----- Original Message ----- > From: "Igor Portnoy" <[EMAIL PROTECTED]> > To: <[EMAIL PROTECTED]> > Sent: Wednesday, June 05, 2002 2:40 PM > Subject: [PHP] Passing variable to new page and pulling the > rest of info > from database > > > Hello, > > > > I am passing a variable to the new page, when user clicks on the link. > Something like that: > > <a href="showimage.php?ID=38"><img src="/some/image.jpg"></a> > > > > How can I extract all other information out of my database for that ID > in the next page (showimage.php)? > > > > Thanks > > > > > > > -- > 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
