Imagick: Convert bash command into php script

Question

There are images I want to remove the background of (or set it to transparent rather). For that reason, I have tested a bash imagick command that looks like this:

convert test.jpg -alpha set -channel RGBA -bordercolor white -border 1x1 -fuzz 2% -fill none -floodfill +0+0 white -shave 1x1 test.png

Because I need to use this in my php script, I need to translate this over now. What I've come up with is this:

$imagick->readImage($path);
$imagick->setImageAlphaChannel(Imagick::ALPHACHANNEL_SET);
$imagick->borderImage('white', 1, 1);
$imagick->floodFillPaintImage('transparent', 20, 'white', 0, 0, false);
$imagick->shaveImage(1, 1);
$imagick->writeImage(str_replace('.jpg', '.png', $path));

From what I can tell, it generates the image and it removes big parts of the background. But the fuzziness setting seems to be ignored.

The results of this script are always the same as when I use -fuzz 0% in the command prompt, no matter what fuzziness value I pass. Am I doing something wrong or is there a bug (which would leave me searching for another script capable of doing this)?

This is not Bash prompt. This is just a normal Bash command. — Arkadiusz Drabczyk
– Arkadiusz Drabczyk, Commented Aug 22, 2015 at 10:57
@thomas good point, I'll do that as a "workaround" / different way that just works. But I'd still like to know why this is so weird. Just out of curiosity :) — Yorrd
– Yorrd, Commented Aug 22, 2015 at 11:05

Danack · Accepted Answer · 2015-08-22 12:43:26Z

1

Am I doing something wrong or is there a bug

Let's call it a documentation error.

Imagick::floodFillPaintImage (and most of the other functions that take a fuzz parameter) need to have the fuzz scaled to the quantum range that ImageMagick was compiled with. e.g. for ImageMagick compiled with 16 bits of depth, the quantum range would be (2^16 - 1) = 65535

There is an example at http://phpimagick.com/Imagick/floodFillPaintImage

$imagick->floodFillPaintImage(
    $fillColor,
    $fuzz * \Imagick::getQuantum(),
    $targetColor,
    $x, $y,
    $inverse,
    $channel
);

So the reason that you're seeing the output image be the same as if you had passed in 0 fuzz, is that ImageMagick is interpreting the value of 2 that you are passing in as 2 / 65535 .... which is approximately zero.

answered Aug 22, 2015 at 12:43

Danack

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1 Comment

Yorrd Over a year ago

damn. That makes sense. Thanks :)

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