Re: [GD] Image errors

On Tue, Jul 14, 2009 at 1:48 PM, Ashley Sheridan<ash@ashleysheridan.co.uk> 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@ashleysheridan.co.uk> wrote: >> > On Tue, 2009-07-14 at 13:27 -0300, Martin Scotta wrote: >> >> $immagine = $tipo($this->updir.$id.'.png'); >> >> >> >> $tipo is undefined >> >> >> >> >> >> On Tue, Jul 14, 2009 at 12:48 PM, Il pinguino >> >> volante<tuxsoul@codeinside.it> wrote: >> >> > Hi to all >> >> > >> >> > I get a problem processing an image with GD libraries. >> >> > >> >> > This is my function >> >> > >> >> >   public function showPicture($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 > > 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.. -- Martin Scotta ps. "tipo" is "type" in Spanish