Hello!!
How are you fetching the rows???
mysql_fetch_rows?? or mysql_fetch_object???
Dan
On Monday, July 8, 2002, at 02:52 PM, [EMAIL PROTECTED] wrote:
> You're right--I did leave out that end quotation mark. Unfortunately,
> though, I'm still getting the same result after adding it.
>
> best,
>
> Matt
>
> ----- Original Message -----
> From: Martin Clifford <[EMAIL PROTECTED]>
> To: <[EMAIL PROTECTED]>; <[EMAIL PROTECTED]>
> Sent: Monday, July 08, 2002 3:47 PM
> Subject: Re: [PHP-DB] passing variables from one page to another
>
>
> If that is your exact code, then you need to have an ending quote after
> the
> query is written. Like this:
>
> $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";
>
> Martin
>
>>>> "Matthew K. Gold" <[EMAIL PROTECTED]> 07/08/02 03:40PM >>>
> 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
> details
> 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, I
> 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
> bgcolor=\"#ff9900\"><th>$CourseID</th></tr><tr><td>$CourseTitle</td></tr>
> </t
> able>");
>
> ?>
>
>
>
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