Ash, Martin,
Seems you are both wandering around the obvious problem...
I suspect that $tipo (in the next line) is *supposed* to be $type - sounds like
a partial Italian translation to me...
So given that $type=imagecreatefrompng (for example, if the mime check returns
'png' - not very
Martin Scotta wrote:
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(
$immagine = $tipo($this-updir.$id.'.png');
$tipo is undefined
On Tue, Jul 14, 2009 at 12:48 PM, Il pinguino
volantetuxs...@codeinside.it wrote:
Hi to all
I get a problem processing an image with GD libraries.
This is my function
public function showPicture($id) {
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
volantetuxs...@codeinside.it wrote:
Hi to all
I get a problem processing an image with GD libraries.
This is my
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
Sheridana...@ashleysheridan.co.uk wrote:
On Tue, 2009-07-14 at 13:27 -0300, Martin Scotta wrote:
$immagine = $tipo($this-updir.$id.'.png');
$tipo is
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
Sheridana...@ashleysheridan.co.uk wrote:
On Tue, 2009-07-14 at 13:27 -0300, Martin Scotta
On Tue, Jul 14, 2009 at 1:48 PM, Ashley
Sheridana...@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
On Tue, 2009-07-14 at 14:16 -0300, Martin Scotta wrote:
On Tue, Jul 14, 2009 at 1:48 PM, Ashley
Sheridana...@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(
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' );
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