Stephan,
Check that there are no html statements or blank lines above the <?php line.
I had a few blank lines in my first program, and it gave me the same error
message.
Hugh
----- Original Message -----
From: Stefan Rusterholz <[EMAIL PROTECTED]>
To: Johan Vikerskog (EMP) <[EMAIL PROTECTED]>;
<[EMAIL PROTECTED]>
Sent: Tuesday, September 25, 2001 12:13 AM
Subject: Re: [PHP] imageJPEG problem.
>
> ----- Original Message -----
> From: "Johan Vikerskog (EMP)" <[EMAIL PROTECTED]>
> To: "'Stefan Rusterholz'" <[EMAIL PROTECTED]>; "Johan Vikerskog
(EMP)"
> <[EMAIL PROTECTED]>; <[EMAIL PROTECTED]>
> Sent: Tuesday, September 25, 2001 9:07 AM
> Subject: RE: [PHP] imageJPEG problem.
>
>
> > Thanks for your answer but the problem remains.
> > It says i allready have given the header information and everything
comes
> out in ascii.
> >
> > file://Johan
> >
> > -----Original Message-----
> > From: Stefan Rusterholz [mailto:[EMAIL PROTECTED]]
> > Sent: den 25 september 2001 08:55
> > To: Johan Vikerskog (EMP); [EMAIL PROTECTED]
> > Subject: Re: [PHP] imageJPEG problem.
> >
> >
> > ----- Original Message -----
> > From: "Johan Vikerskog (EMP)" <[EMAIL PROTECTED]>
> > To: <[EMAIL PROTECTED]>
> > Sent: Tuesday, September 25, 2001 8:47 AM
> > Subject: [PHP] imageJPEG problem.
> >
> >
> > > When trying to use a simple code like this below i receive an error
> which
> > says that it cant find the header.
> > >
> > > <?
> > > Header("Content-type: image/jpeg");
> >
> > I don't know about the other stuff of your script, but when I wrote a
> > similar script, I took "content-type: image/jpg" and not 'jpeg'.
> > Probably that's the point...
> >
> > > $im = imagecreatefromjpeg("./test.jpg");
> > > Imagejpeg($im,'',20);
> > > ImageDestroy($im);
> > > ?>
> > >
> > > It outputs the information in ascii and binary code. it doesn't show
the
> > picture
> > >
> > > Thanks for all the help you can give me.
> > > file://Johan
> > >
> > > --
> > > PHP General Mailing List (http://www.php.net/)
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>
> make sure you run the php-part at the very top of the file:
> <?php
> header(...);
> ?>
> <HTML>
> <HEAD>
> ...and so on
>
>
> if you have output before you execute the php part, your webserver will
> generate the header for you and then it is impossible to send another
header
> (exactly because of what the warning tells you: the headers are already
> sent - but by the server and not by you and the server normally assumes
> text/html or text/plain).
>
> I hope, it helps
> regards
>
> Stefan Rusterholz
>
>
>
>
>
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