I am trying to create a PHP script that takes an image:
http://i.stack.imgur.com/eNvlM.png
and then applies a PNG image:
http://i.stack.imgur.com/iJr2I.png
as a mask.
The end result needs to maintain transparency:
http://i.stack.imgur.com/u0l0I.png
If at all possible I want to do this in GD, ImageMagick is not really an option right now.
How would I go about this?
phalacee's post (in "PHP/GD, how to copy a circle from one image to another?") seems to be along the right lines but I specifically need to use an image as a mask, not a shape.
Matt,
If you make your png with the oval white fill on black background instead of black fill with transparent background the following function does it.
<?php // Load source and mask $source = imagecreatefrompng( '1.png' ); $mask = imagecreatefrompng( '2.png' ); // Apply mask to source imagealphamask( $source, $mask ); // Output header( "Content-type: image/png"); imagepng( $source ); function imagealphamask( &$picture, $mask ) { // Get sizes and set up new picture $xSize = imagesx( $picture ); $ySize = imagesy( $picture ); $newPicture = imagecreatetruecolor( $xSize, $ySize ); imagesavealpha( $newPicture, true ); imagefill( $newPicture, 0, 0, imagecolorallocatealpha( $newPicture, 0, 0, 0, 127 ) ); // Resize mask if necessary if( $xSize != imagesx( $mask ) || $ySize != imagesy( $mask ) ) { $tempPic = imagecreatetruecolor( $xSize, $ySize ); imagecopyresampled( $tempPic, $mask, 0, 0, 0, 0, $xSize, $ySize, imagesx( $mask ), imagesy( $mask ) ); imagedestroy( $mask ); $mask = $tempPic; } // Perform pixel-based alpha map application for( $x = 0; $x < $xSize; $x++ ) { for( $y = 0; $y < $ySize; $y++ ) { $alpha = imagecolorsforindex( $mask, imagecolorat( $mask, $x, $y ) ); $alpha = 127 - floor( $alpha[ 'red' ] / 2 ); $color = imagecolorsforindex( $picture, imagecolorat( $picture, $x, $y ) ); imagesetpixel( $newPicture, $x, $y, imagecolorallocatealpha( $newPicture, $color[ 'red' ], $color[ 'green' ], $color[ 'blue' ], $alpha ) ); } } // Copy back to original picture imagedestroy( $picture ); $picture = $newPicture; } ?>
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