On Tue, 2009-07-14 at 13:41 -0300, Martin Scotta wrote:
> He is calling the function by variable
>
> something like this
>
> $func = 'var_dump';
>
> $func( new Foo );
>
> On Tue, Jul 14, 2009 at 1:30 PM, Ashley
> Sheridan<a...@ashleysheridan.co.uk> wrote:
> > On Tue, 2009-07-14 at 13:27 -0300, Martin Scotta wrote:
> >> $immagine = $tipo($this->updir.$id.'.png');
> >>
> >> $tipo is undefined
> >>
> >>
> >> On Tue, Jul 14, 2009 at 12:48 PM, Il pinguino
> >> volante<tuxs...@codeinside.it> wrote:
> >> > Hi to all
> >> >
> >> > I get a problem processing an image with GD libraries.
> >> >
> >> > This is my function
> >> >
> >> > public function showPicture($id) {
> >> > header('Content-type: image/jpeg');
> >> > $mime = mime_content_type($this->updir.$id.'.png');
> >> > $type = explode('/',$mime);
> >> > $type = 'imagecreatefrom'.$type[1];
> >> > $immagine = $tipo($this->updir.$id.'.png');
> >> > imagejpeg($immagine,null,100);
> >> > imagedestroy($immagine);
> >> > }
> >> >
> >> > If i commentize the "header" function i get a lot of strange simbols
> >> > (such
> >> > as the code of a jpeg image!) but no errors.
> >> > The result of this code is a blank page. In Firefox the title sets to
> >> > "picture.php (JPEG image)" and if i right-click it and click on
> >> > "Proprieties" i get "0px × 0px (resized as 315px × 19px)".
> >> >
> >> > P.S.: I get the same result when I write $immagine =
> >> > imagecreatefromjpeg(...)
> >> >
> >> > (Sorry for my english)
> >> >
> >> > Thanks in advance,
> >> > Alfio.
> >> >
> >> > --
> >> > PHP General Mailing List (http://www.php.net/)
> >> > To unsubscribe, visit: http://www.php.net/unsub.php
> >> >
> >> >
> >>
> >>
> >>
> >> --
> >> Martin Scotta
> >>
> > Also, it doesn't look like you're actually doing anything with $type
> >
> > Thanks
> > Ash
> > www.ashleysheridan.co.uk
> >
> >
>
>
>
> --
> Martin Scotta
>
Bottom post ;)
$type = explode('/',$mime);
$type = 'imagecreatefrom'.$type[1];
$immagine = $tipo($this->updir.$id.'.png');
imagejpeg($immagine,null,100);
imagedestroy($immagine);
I'm not sure you understood what I meant. line 2 above $type is assigned
to the string 'imagecreatefrom'.$type[1];
Now I assume that was to be used later in some sort of eval() statement,
but its never called again, so the line really does nothing.
Thanks
Ash
www.ashleysheridan.co.uk
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