ID:               21763
 User updated by:  [EMAIL PROTECTED]
 Reported By:      [EMAIL PROTECTED]
 Status:           Bogus
 Bug Type:         GD related
 Operating System: Mac OS X 10.2.3
 PHP Version:      4.3.0
 New Comment:

Actually, this particular case was a simplification of what I'm
eventually trying to accomplish. What I really want is the black in any
image to be transparent when the image is copied, and I cannot
guarantee that a given pixel (such as 0,0) will be black among the set
of images I intend to work with.

As for the documentation, it states that "imagecolorallocate() returns
a color identifier representing the color composed of the given RGB
components". If this is to be used _only_ for retrieving the index of
_new_ colors, the documentation should say as much. (What the
"allocate" part means is ambiguous. What actually gets allocated, the
name or the color? Could be either or both.)

It sounds like what I really want to do is this:

imagecolortransparent($image2, imagecolorexact($image2, 0, 0, 0) );

But what happens if both blacks (index values 2 and 3) are used? I
suppose this requires either some weird and/or thorough manipulations
of the image and its colors, or that these functions consider the
possibility that there may be more than one index with a given color.


Previous Comments:
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[2003-01-20 17:06:30] [EMAIL PROTECTED]

Thank you for taking the time to write to us, but this is not
a bug. Please double-check the documentation available at
http://www.php.net/manual/ and the instructions on how to report
a bug at http://bugs.php.net/how-to-report.php

You are making an existing color transparent therefor you don't need to
do ImageColorAllocate(). Simply do imagecolorat() on the pixel with the
color you wish to make transparent and use the return value as the 2nd
argument to imagecolortransparent().
In your particular case this can be done with:
$black = ImageColorAt($image2, 0, 0);

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[2003-01-20 15:51:18] [EMAIL PROTECTED]

When you give feedback, please DO NOT USE the page behind the 'Add
Comment' link!!


------------------------------------------------------------------------

[2003-01-20 12:20:57] [EMAIL PROTECTED]

I'm using the bundled library. And since there seems to be no way to
upload them (even though they're about 5K total), I'll just put the
images here:

Orc: http://homepage.mac.com/tuxedobob/Orc.png
Board: http://homepage.mac.com/tuxedobob/Board.png

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[2003-01-20 09:39:44] [EMAIL PROTECTED]

Please provide the two images used in your example.
Also, are you using the bundled GD library or the non-bundled one?

------------------------------------------------------------------------

[2003-01-20 02:28:08] [EMAIL PROTECTED]

In the following script:

<?php
$image2 = ImageCreateFromPNG('images/Monsters/Orc.png');
$image = ImageCreateFromPNG('images/Board.png');
$black = ImageColorAllocate($image2, 0, 0, 0);

imagecolortransparent($image2, $black);
imagecopy($image, $image2, 40*2+8, 40*2+8, 0, 0, 39, 39);

header("Content-type: image/png");
ImagePNG($image);
imagedestroy($image);
imagedestroy($image2);
?>

what I expect to happen is that the black in the 'Orc.png' image will
become transparent when it is copied onto 'Board.png'. This doesn't
happen. The transparency is not recognized at all.

More info:
Orc.png is a 2-bit PNG, with black occupying two of the colors. (It
doesn't work even if it only has one.)
Board.png is a 4-bit PNG with 6 distinct colors (11 copies of black).
If I change the line to

imagecolortransparent('$image2, imagecolorat($image2, 0, 0);

it works correctly. If I change the line use either 2 or 3, as in

imagecolortransparent('$image2, 3);

it works depending on which black (index no. 2 or 3) is actually in the
image. There may be no way around this except to have
imagecolortransparent check by RGB values rather than an index number.

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