PHP GD：将图像颜色乘以色调 [英] PHP GD: Multiply image colors with tint color

问题描述

我可以使用GD将给定颜色（RGB）乘以图像资源（RGBA）的每个像素吗？例如，如果给定的色调颜色是红色（255,0,0），则图像资源上的黑色像素应该保持黑色（因为255 * 0 = 0），并且更亮的像素应该受到更多的红色色调因素的影响。 / p>

我尝试了

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

但只更改黑色像素。 NEGATE，IMG_FILTER_COLORIZE，NEGATE也不会工作。

解决方案

这似乎对我有用：

 <？php 
 //用于从任何上传的文件创建图像的函数
 
 function imageCreateFromAny（$ filepath）{
 
 $ type = exif_imagetype（$ filepath）; // []如果没有exif，你可以使用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; 
} 
 
 //设置变量
 $ filter_r = 255; 
 $ filter_g = 0; 
 $ filter_b = 0; 
 $ suffixe = -  red; 
 $ path =original-source / image.jpg; 
 
 if（is_file（$ path））{
 $ image = imageCreateFromAny（$ path）; 
 
 //反转inteded颜色red
 $ filter_r_opp = 255  -  $ filter_r; // = 0 
 $ filter_g_opp = 255  -  $ filter_g; // = 255 
 $ filter_b_opp = 255  -  $ filter_b; // = 255 
 // color is nowaqua
 
 / * 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'红色阴影失败。 
} 
 
？> 
  

can I use GD to multiply given color (RGB) with every pixel of an image resource (RGBA)? For example, if the given tint color is red (255, 0, 0), a black pixel on the image resource should stay black (since 255 * 0 = 0) and brighter pixels should be affected more by that red tint factor.

I tried

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

but that only changes black pixels. NEGATE, IMG_FILTER_COLORIZE, NEGATE also won't work.

解决方案

This seems to work for me:

<?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.';
}

?>

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