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
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
