Php Image Transformation not Doing What I Expect It to Do - php

Question

The following block of php code is supposed to take this image as input snd produce this image as output (convert the black to yellow and the light-blue to black):
However, I'm getting this image as output instead.
Can anyone see the problem with my code?
$im = imagecreatefrompng("./input.png");
$width = imagesx($im);
$height = imagesy($im);
$new = imagecreate($width, $height);
imagecopy($new, $im, 0, 0, 0, 0, $width, $height);
imagecolorset($new, imagecolorexact($new, 0, 0, 0), 255, 255, 0);
for($i = 0; $i < $width; $i++) {
for($j = 0; $j < $height; $j++) {
$index = imagecolorat($new, $i, $j);
$rgb = imagecolorsforindex($new, $index);
if($rgb['red'] != 255 && $rgb['green'] != 255 && $rgb['blue'] != 0) {
echo '(' . $i . ', ' . $j . ')' . 'color => (' . (255 - $rgb['blue']) . ', ' . (255 - $rgb['blue']) . ', 0)<br />';
$color = imagecolorallocate($new, 255 - $rgb['blue'], 255 - $rgb['blue'], 0);
imagesetpixel($new, $i, $j, $color);
}
unset($index);
unset($rgb);
}
}
imagepng($new, 'tesst.png');
imagedestroy($im);
imagedestroy($new);

Answer 1

I believe the source of the issue here is that when using a palette based image, such as the one you have created by calling imagecreate(), it is possible to declare the same color at multiple indexes within the palette.
This means that, because you are calling imagecolorallocate() on every iteration, the palette eventually becomes full and imagecolorallocate() starts returning false (this can be seen if you var_dump($color); before the imagesetpixel() call). false evaluates to zero when cast to an integer - so when the palette fills up, all remaining pixels are effectively converted to the background color.
There are two things that you could do about this, the first and probably easiest is just to use a true-color image, which is just a simple case of changing imagecreate($width, $height); to imagecreatetruecolor($width, $height);.
If you want to stick with the palette-based image (for example for reasons of output image data size - with an image containing so few colors, a palette-based image will be considerably smaller), you will need to cache the allocated colors manually so you can re-use them, something like this:
// ...
$colors = array();
for ($x = 0; $x < $width; $x++) { // iterate x axis
for ($y = 0; $y < $height; $y++) { // iterate y axis
// Get the color at this index
$index = imagecolorat($new, $x, $y);
// Only allocate a new color if not already done
if (!isset($colors[$index])) {
$rgb = imagecolorsforindex($new, $index);
if ($rgb['red'] != 255 || $rgb['green'] != 255 || $rgb['blue'] != 0) {
// If it's not the background color allocate a new color
$r = $g = 255 - $rgb['blue'];
$b = 0;
$colors[$index] = imagecolorallocate($new, $r, $g, $b);
} else {
// Otherwise set the index to false, we can ignore it
$colors[$index] = false;
}
}
// If there's something to do, do it
if ($colors[$index] !== false) {
imagesetpixel($new, $x, $y, $colors[$index]);
}
}
}
// ...
You may also wish to track the colors in use in the image so you can "cleanse the palette" afterwards (i.e. deallocate any colors that are no longer in use in the image, which will further help with the data size reduction). Although arguably it would be better, in this case, to start with a clean palette and inspect the old image resource to get the pixel details, instead of copying the original to a new image resource.

Answer 2

yes your
$color = imagecolorallocate($new, 255 - $rgb['blue'], 255 - $rgb['blue'], 0);
is messing everything up..
if you desire the same output... just paste the line outside for loop this will solve your problem if you want the specific image:
$color = imagecolorallocate($new, 35, 35, 0); //got from debugging
and it will get u the desired output.
Dins