On Tue, 2009-07-14 at 14:16 -0300, Martin Scotta wrote:
> On Tue, Jul 14, 2009 at 1:48 PM, Ashley
> Sheridan<ash@xxxxxxxxxxxxxxxxxxxx> wrote:
> > 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<ash@xxxxxxxxxxxxxxxxxxxx> wrote:
> >> > On Tue, 2009-07-14 at 13:27 -0300, Martin Scotta wrote:
> >> >> $immagine = $tipo($this->updir.$id.'.png');
> >> >>
> >> >> $tipo is undefined
> >> >>
> >> >>
> >> >> On Tue, Jul 14, 2009 at 12:48 PM, Il pinguino
> >> >> volante<tuxsoul@xxxxxxxxxxxxx> wrote:
> >> >> > Hi to all
> >> >> >
> >> >> > I get a problem processing an image with GD libraries.
> >> >> >
> >> >> > This is my function
> >> >> >
> >> >> > public function showPicture($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
> >
> >
>
> Mmmm, No
>
> $type = explode('/',$mime); # type is array
> $type = 'imagecreatefrom'.$type[1]; # type is a string
>
> He is actually re-assigning the var ussing the content of the var.
> Sounds crazy, but I use this method a lot, it helps to keep the scope clean.
>
It's not crazy, I do it a lot, but he *does nothing with it* after that,
and the scope of the variable is limited to the function.
Thanks
Ash
www.ashleysheridan.co.uk
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
