Actually, I am kind of confused.. =P ...why wouldn't SELECT * FROM table
WHERE (artist_id = '$id') work?
--
David Balatero
[EMAIL PROTECTED]
--
-Original Message-
From: Matt Nigh [mailto:[EMAIL PROTECTED]]
Sent: Sunday, July 22, 2001 6:30 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] php/my
hi. i was wondering if the knowledgeable people on this list could help me out with a
problem i'm having.
what i need is this:
i want a single page that with a variable taken from a database, will display a
different page depending on the name of a band.
http://www.site.com/bands.php?artist=theb
Hi All:
I have a problem. I trying to create a temporary table from the query
results of the following statement.
$results=mysql_query ("SELECT CONCAT(membername, \" \", mission, \"
\", brief, \" \", city, \" \", state, \" \", programs) AS searchable,
memberid FROM members") or die("DIE! DIE!
Hi Tony
Don't know if you got any feedback on this, but I would crate a form on the
HTML page with an input type (say a dropdown list) with the name of, for
example, itemnumber.
The code could be something like this...
Item description
Item description
Item description
etc
then in inventor
When you execute your query, do like I do...
$query = mysql_db_query($db, $query) or die(mysql_error());
the "or die(mysql_error())" will tell you what the mysql error is...
On Sun, 22 Jul 2001, Darrell wrote:
> I got mysql installed on my win2000 machine, and have apache running with
> php 4
I got mysql installed on my win2000 machine, and have apache running with
php 4 all working correctly.
but I can't get php to work with the db...
I'm pretty sure I can connect with mysql_connect(...) because it doesn't die
at that spot. rather, when I try to do the mysql_query(...) it dies each