ID: 21763 Updated by: [EMAIL PROTECTED] Reported By: [EMAIL PROTECTED] -Status: Open +Status: Bogus Bug Type: GD related Operating System: Mac OS X 10.2.3 PHP Version: 4.3.0 New Comment:
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 Then you write a function to generate you a color list of the image and use that. This is NOT a PHP bug. Previous Comments: ------------------------------------------------------------------------ [2003-01-20 18:32:42] [EMAIL PROTECTED] Oops. I think I should have changed this back to "open". ------------------------------------------------------------------------ [2003-01-20 18:27:06] [EMAIL PROTECTED] 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. ------------------------------------------------------------------------ [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); ------------------------------------------------------------------------ [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 ------------------------------------------------------------------------ The remainder of the comments for this report are too long. To view the rest of the comments, please view the bug report online at http://bugs.php.net/21763 -- Edit this bug report at http://bugs.php.net/?id=21763&edit=1
