Re: [PHP] [GD] Image errors
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 reliable, I suspect),
then
$immagine = $type($this-updir.$id.'.png')
should create a GD resource from the file, but the image appears to be empty.
My take on this is:
OP says he gets the same result from
$immagine = imagecreatefromjpeg(this-updir.$id.'.png')
- well I might expect to get an error message if I loaded a PNG expecting it to
be a JPEG, but I certainly wouldn't expect an image.
On some basic tests, I find that mime_content_type() is not defined on my
system, so ignoring that and trying to load a PNG file using imagecreatefromjpeg
results in pretty much the same result as the OP...
Conclusions:
First: if you use Italian for your variable names, don't change half of their
instances to English...
Second: Make sure you actually know the mime type of a file before you try to
load it into GD...
My version of this would test against known image types to try the GD function:
foreach (Array('png','jpeg','gif') as $typeName)
{
$type = 'imagecreatefrom'.$typeName;
$immagine = $type(this-updir.$id.'.png'le);
if (is_resource($immagine))
{
header('Content-type: image/jpeg');
imagejpeg($immagine,null,100);
imagedestroy($immagine);
break;
}
}
header('HTTP/1.0 500 File is not an allowed image type');
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Developer fax: 01580 893399
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Re: [PHP] [GD] Image errors
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( $this-updir . $id . '.png' );
}
hey, look, just 2 lines!
But it doesn't convert the image from whatever came in to a JPEG output, which
is what the OP's code appears to be trying to do (and possibly ought to work...)
--
Peter Ford phone: 01580 89
Developer fax: 01580 893399
Justcroft International Ltd., Staplehurst, Kent
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Re: [PHP] [GD] Image errors
$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) {
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.
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Re: [PHP] [GD] Image errors
On Tue, 2009-07-14 at 13:27 -0300, Martin Scotta wrote:
$immagine = $tipo($this-updir.$id.'.png');
$tipo is undefined
On Tue, Jul 14, 2009 at 12:48 PM, Il pinguino
volantetuxs...@codeinside.it wrote:
Hi to all
I get a problem processing an image with GD libraries.
This is my function
public function showPicture($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.
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--
Martin Scotta
Also, it doesn't look like you're actually doing anything with $type
Thanks
Ash
www.ashleysheridan.co.uk
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Re: [PHP] [GD] Image errors
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');
$tipo is undefined
On Tue, Jul 14, 2009 at 12:48 PM, Il pinguino
volantetuxs...@codeinside.it wrote:
Hi to all
I get a problem processing an image with GD libraries.
This is my function
public function showPicture($id) {
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.
--
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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
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Re: [PHP] [GD] Image errors
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 wrote:
$immagine = $tipo($this-updir.$id.'.png');
$tipo is undefined
On Tue, Jul 14, 2009 at 12:48 PM, Il pinguino
volantetuxs...@codeinside.it wrote:
Hi to all
I get a problem processing an image with GD libraries.
This is my function
public function showPicture($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
--
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Re: [PHP] [GD] Image errors
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
Sheridana...@ashleysheridan.co.uk wrote:
On Tue, 2009-07-14 at 13:27 -0300, Martin Scotta wrote:
$immagine = $tipo($this-updir.$id.'.png');
$tipo is undefined
On Tue, Jul 14, 2009 at 12:48 PM, Il pinguino
volantetuxs...@codeinside.it wrote:
Hi to all
I get a problem processing an image with GD libraries.
This is my function
public function showPicture($id) {
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
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Re: [PHP] [GD] Image errors
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( 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');
$tipo is undefined
On Tue, Jul 14, 2009 at 12:48 PM, Il pinguino
volantetuxs...@codeinside.it wrote:
Hi to all
I get a problem processing an image with GD libraries.
This is my function
public function showPicture($id) {
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
Re: [PHP] [GD] Image errors
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
Sheridana...@ashleysheridan.co.uk wrote:
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( 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');
$tipo is undefined
On Tue, Jul 14, 2009 at 12:48 PM, Il pinguino
volantetuxs...@codeinside.it wrote:
Hi to all
I get a problem processing an image with GD libraries.
This is my function
public function showPicture($id) {
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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