Why are you ussing GD?
All you need is output the image to the browser?
try this... I didn't test/run this code, but it may work...
public function showPicture( $id ) {
header('Content-type:' . mime_content_type( $this->updir . $id .
'.png' ) );
readfile( $this->updir . $id . '.png' );
}
hey, look, just 2 lines!
On Tue, Jul 14, 2009 at 2:20 PM, Ashley
Sheridan<ash@xxxxxxxxxxxxxxxxxxxx> wrote:
> 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
>
>
--
Martin Scotta
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