I am getting link from visitor using form, that means I don't have control
what they would type. e.g they may type
http://www.suggestedlink.com/myfav.wav, www.suggestedlink.com/myfav.ram and
http://suggestedlink.com/myfav.mp3  or so I am looking general version which
can handle all.

and on top of that I wanted to play music file in to the default player, but
it did not work. I used something like this.

select music_file from music where id=\"$id\"
$temp = $result
so everything is working fine except I got file not found, and the file path
is something like this  http://www.mydomain.com/www.suggestedlink/myfav.mp3
so what should I do.

On 12/7/05, Edward Gray <[EMAIL PROTECTED]> wrote:
> actually, you may want to check the source of the page.  if the url does
> not start with "http://"; (or https, or ftp, etc.), browsers will assume the
> link is on the current server.  how are you storing the urls?  as full urls,
> as domain/path/file.htm, ....?  if all of your urls should start with
> "http://";, you could either store them in the database that way or
> echo 'http://' . $temp;
> hope this helps.
> Edward Gray
> Web Development Team
> University of Mary Washington
> 540-654-1564
> >>> "Bastien Koert" <[EMAIL PROTECTED]> 12/06/05 1:44 PM >>>
> $url = str_replace("www.mydomain.com","",$url)
> bastien
> >From: Mohamed Yusuf <[EMAIL PROTECTED]>
> >To: php-db@lists.php.net
> >Subject: [PHP-DB] problem of retrieving urls from mysql
> >Date: Tue, 6 Dec 2005 09:51:18 -0800
> >
> >I would like to store and retrieve urls, but I have problem which is I
> get
> >my url + the other url, Instead I should get another url only
> >
> >$temp = linkurl;
> >echo "$temp";
> >
> >the echo prints something like this
> >http://www.mydomain.com/www.otherlink.com
> >
> >so I wan get rid off my url and get only other link, so how can I do
> that?
> >
> >thanks in advance
> --
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