Hmm, I am still getting a parse error on the last line of code...
In a message dated 12/16/2002 1:49:26 PM Eastern Standard Time,
[EMAIL PROTECTED] writes:
> $sql = "SELECT .....";
> $sql_e = mysql_query($sql);
>
> while ($result = mysql_fetch_array($query_e)) {
> .....
> }
>
> You were missing the mysql_query
>
>
> <[EMAIL PROTECTED]> escribió en el mensaje
> [EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> >I am trying to display a column from my database as a list. Each listing
> >needs to be a URL that links to another script that brings up all of the
> data
> >in the row to edit. I keep getting a parser error and I can't figure it
> out.
> > Here is the code and any help is greatly appreciated.
> >
> ><?php
> >$db = mysql_connect("localhost", "Uname", "PW");
> >
> >$select_db = mysql_select_db("machmedi_meetingRequest",$db);
> >$sql = "SELECT * FROM requests";
> >
> >while ($result = mysql_fetch_array($query)) {
> >$id= $result["id"];
> >$meetingName= $result["meetingName"];
> >
> >echo ("<A HREF="/"edit.php?id='$id'\"">$meetingName</A> ");
> >?>
> >
>
>
Christopher Parker
Senior Corporate Technical Specialist, Corporate Events
America Online Inc.
Phone: 703.265.5553
Fax: 703.265.2007
Cell: 703.593.3199
