[PHP] Passing variable to new page and pulling the rest of info from database

2002-06-05 Thread Igor Portnoy
Hello, I am passing a variable to the new page, when user clicks on the link. Something like that: a href=showimage.php?ID=38img 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

Re: [PHP] Passing variable to new page and pulling the rest of info from database

2002-06-05 Thread 1LT John W. Holmes
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

RE: [PHP] Passing variable to new page and pulling the rest of info from database

2002-06-05 Thread Scott Hurring
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:

Re: [PHP] Passing variable to new page and pulling the rest of info from database

2002-06-05 Thread 1LT John W. Holmes
Or simply just validate it's an integer, like it should be, like you would do with any user input... There is no mysql_quote() function...or am I missing something? ---John Holmes... - Original Message - From: Scott Hurring [EMAIL PROTECTED] To: Php-General (E-mail) [EMAIL PROTECTED]

RE: [PHP] Passing variable to new page and pulling the rest of info from database

2002-06-05 Thread Scott Hurring
Sorry, i meant: mysql_escape_string( $value ) I use a class to handle accessing the database, and i named the function $db-quote(...); i keep forgetting that the actual function call that the class is making ;-) And yes, validating that it's an integer works also, but even after pregging vars