I think you can get this error when you sql query fails.
Try copy and pasting the $sql query to the mysql command line and see what
response you get (it looks as though this would work though unless you
haven't got a table called test).
I guess it might be that you are not successfully logging on
Can anyone spot why I get the error
Warning: Failed opening
'd:/www/htdocs/ski-info-online/poll.php?id=KnChooseBoots' for inclusion
(include_path='') in d:\WWW\htdocs/ski-info-online/skiKnSkiBoots.php on line
113
Here is the calling line:
?php include($DOCUMENT_ROOT .
err...what javascript the form is HTML?
Try...
form name=NAME method=get OR post action=?php echo $PHP_SELF?
Then in your page have a php tag that checks for the presence of $Find (in
your case), e.g.
if ($Find) {
sql = (SELECT * FROM table WHERE $criteria = value);
then the rest...
}
lesystem?
Peter
-Original Message-
From: TorrentUK [mailto:[EMAIL PROTECTED]]
Sent: 13 November 2001 12:47
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Problem with include()
Can anyone spot why I get the error
Warning: Failed opening
'd:/www/htdocs/ski-info-online/pol
Original Message -
From: TorrentUK [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Tuesday, November 13, 2001 11:24 AM
Subject: Re: [PHP-DB] Problem with include()
Hi Peter,
I'll be surprised if that's it as I am running this through an Apache
server
which insists on the \ being /
--
From: TorrentUK [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Tuesday, November 13, 2001 1:16 PM
Subject: Re: [PHP-DB] Problem with include()
According to the php manual it can pass args and values.
The value I pass in through the $id then determines what poll is
presented
to the
Please could some take a look at this code and tell me why when I take my IF
statements out of the function and put in them in the same place where I
call the function from they work, but as soon as I replace them with the
function name they don't?
Appreciate any help.
torrent
Here's the
Yep, that'll be it.
Thanks for the help. :)
torrent
Charles F. McKnight [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
On Tue, 30 Oct 2001, TorrentUK wrote:
SNIP
you either need to pass the variables to the function or use global
v
= select * from tbl where ctry='$country[0]';
for ($n=1; count($country) $n; $n++) {
$sql.= or ctry='$country[$n]';
}
richard
TorrentUK wrote:
I am designing a ski web site and am presently trying to put together a
resort database. I started the search script tonight and am quite
I am designing a ski web site and am presently trying to put together a
resort database. I started the search script tonight and am quite pleased
that it works. Where I am struggling is that I would like my visitors to be
able to select multiple countries from the drop down list and for my query
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