Question

我可以使用GD将给定颜色（RGB）与图像资源（RGBA）的每个像素相乘吗？例如，如果给定的色调颜色为红色（255,0,0），则图像资源上的黑色像素应保持黑色（因为255 * 0 = 0），较亮的像素应受该红色色调因子的影响。 / p>

我试过

imagefilter($sprite, IMG_FILTER_COLORIZE, 255, 0, 0);

但这只会改变黑色像素。 NEGATE，IMG_FILTER_COLORIZE，NEGATE也没有工作。

Answer 1

这似乎对我有用：

<?php
// function for creating images from any uploaded

function imageCreateFromAny($filepath) { 

    $type = exif_imagetype($filepath); // [] if you don't have exif you could use getImageSize() 
    $allowedTypes = array( 
        1,  // [] gif 
        2,  // [] jpg 
        3,  // [] png 
        6   // [] bmp 
    ); 
    if (!in_array($type, $allowedTypes)) { 
    return false; 
    } 
    switch ($type) { 
        case 1 : 
            $im = imageCreateFromGif($filepath); 
        break; 
        case 2 : 
            $im = imageCreateFromJpeg($filepath); 
        break; 
        case 3 : 
            $im = imageCreateFromPng($filepath); 
        break; 
        case 6 : 
            $im = imageCreateFromBmp($filepath); 
        break; 
    }    
    return $im;  
}

// set up variables
$filter_r=255;
$filter_g=0;
$filter_b=0;
$suffixe="-red";
$path="original-source/image.jpg";

if(is_file($path)){
    $image=imageCreateFromAny($path);

    // invert inteded color "red"
    $filter_r_opp = 255 - $filter_r; // = 0
    $filter_g_opp = 255 - $filter_g; // = 255
    $filter_b_opp = 255 - $filter_b; // = 255
    // color is now "aqua"

    /* FAST METHOD */
    imagefilter($image, IMG_FILTER_NEGATE);
    imagefilter($image, IMG_FILTER_COLORIZE, $filter_r_opp, $filter_g_opp, $filter_b_opp);
    imagefilter($image, IMG_FILTER_NEGATE);

    $new_path=substr($path,0,strlen($path)-4).$suffixe.".jpg";
    imagejpeg($image,$new_path);

    imagedestroy($image);

    echo 'Red shading success.';
}
else {
    echo 'Red shading failed.';
}

?>

PHP GD：将图像颜色与色调相乘

1 个答案: