Hi, I would like to test you code in my webpage, but
i dont really know how to get along with path informations
like ../images/pictures/image24.jpg
Do I have to split the string in multiple parts, so i at last only
got the picture name? But how then do i open the file?
- thru fopen() or something equal or how do you do it?
Please gimme some more informations if you like :))
Greetings
Sascha
----- Original Message -----
From: "Ernest E Vogelsinger" <[EMAIL PROTECTED]>
To: "Dara Dowd" <[EMAIL PROTECTED]>
Cc: <[EMAIL PROTECTED]>; <[EMAIL PROTECTED]>
Sent: Thursday, January 30, 2003 10:27 AM
Subject: Re: [PHP] content type header for downloading files
> At 02:04 30.01.2003, Dara Dowd said:
> --------------------[snip]--------------------
> >$fname is the name of the file i want to download. It works in IE, as in
the
> >name of the file to be downloaded appears correctly in the dialog box. I
> >changed 'application/octet' to 'application/octet-stream' and this had no
> >effect.
> --------------------[snip]--------------------
>
> This is what I do to transmit a file for a save-as download, and it works
> in all browsers I know and have tested the same (Netscape, IE, Opera/Win;
> Mozilla, Konqueror et al. / Linux):
>
> header('Content-type: application/octet-stream');
> header('Content-length: ' . strlen($this->export));
> header('Content-Disposition: attachment' .
> (empty($fname) ? '' : ';filename="'.$fname.'.csv"'));
> echo $this->export;
> exit();
>
> Note that it seems to be necessary to surround the filename using quotes
to
> make it work correctly.
>
>
>
> --
> >O Ernest E. Vogelsinger
> (\) ICQ #13394035
> ^ http://www.vogelsinger.at/
>
>
>
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