It works from the command line if I leave out course.CourseID =  $CourseID
in the where line of the select clause.  I assume that I'd have to leave
that out when working from the command line because it's defined in the php

But I think that that is where my problem is.  As I've mentioned, I'm very
new at this, so the answer could be very basic.



----- Original Message -----
From: Hutchins, Richard <[EMAIL PROTECTED]>
Sent: Monday, July 08, 2002 4:35 PM
Subject: RE: [PHP-DB] passing variables from one page to another

> Matthew, have you tried running your query from the command line in MySQL?
> If you can do that successfully, that'll prove that the query is
> properly and the problem can be tracked elsewhere.
> -----Original Message-----
> From: Matthew K. Gold [mailto:[EMAIL PROTECTED]]
> Sent: Monday, July 08, 2002 3:40 PM
> Subject: [PHP-DB] passing variables from one page to another
> Hi Everybody,
> Please forgive the basic nature of this question--I have looked at the
> manual, but I found that much of it went over my head.
> I'm trying to create a second level of a website that displays course
> listings.  The first level lists a bunch of courses.  I'd like users to be
> able to click on the title of a course to go to a page that contains
> about that course.
> From what I understand, I should do this by making the title of the course
> on the first page (courses.php) a link to the second page (courseinfo.php)
> with a query string attached--so that the link would look like
>  <a href="courseinfo.php?CourseID=12">Accounting 101</a>
> What I'm having trouble doing is coding the second page.  Here's what I've
> done.  When I run this, I get a blank page in return...if anyone can help,
> would greatly appreciate it.  Thanks in advance.
> Matt
> <?php
> $db = @mysql_connect("host","user","pword");
> mysql_select_db("dln", $db);
> if ( !$CourseID ) {
> print ("no course id included")
> exit;
> };
> $query = "Select * from course, disc, instit, prof where course.CourseID =
> $CourseID and course.DiscID = disc.DiscID and course.InstitID =
> instit.InstitID and course.ProfID = prof.ProfID;
> $result = mysql_query ($query);
> echo "MySQL error number " . mysql_errno() . ": " . mysql_error();
> print ("<table border=\"0\" cellpadding=\"5\" class=\"default\">\n");
> print ("<tr
> able>");
> ?>
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