如何在 PHP 中绘制稍微对角的渐变填充？

Question

我发现下面的函数可以在PHP中绘制垂直渐变。然而，许多网页设计师喜欢他们的渐变有一个左上角的光源，以使渐变看起来更真实。那么，如何在垂直渐变上稍微改变角度，使其变成轻微渐变呢？我不想完全做得太过分，只是当它沿着垂直梯度向下移动时，只需向右轻微移动即可。

<?php

function hex2rgb($sColor) {
    $sColor = str_replace('#','',$sColor);
    $nLen = strlen($sColor) / 3;
    $anRGB = array();
    $anRGB[]=hexdec(str_repeat(substr($sColor,0,$nLen),2/$nLen));
    $anRGB[]=hexdec(str_repeat(substr($sColor,$nLen,$nLen),2/$nLen));
    $anRGB[]=hexdec(str_repeat(substr($sColor,2*$nLen,$nLen),2/$nLen));
    return $anRGB;
}

$nWidth = 960;
$nHeight = 250;
$sStartColor = '#2b8ae1';
$sEndColor = '#0054a1';
$nStep = 1;

$hImage = imagecreatetruecolor($nWidth,$nHeight);
$nRows = imagesy($hImage);
$nCols = imagesx($hImage);
list($r1,$g1,$b1) = hex2rgb($sStartColor);
list($r2,$g2,$b2) = hex2rgb($sEndColor);
$nOld_r = 0; $nOld_g = 0; $nOld_b = 0;
for ( $i = 0; $i < $nRows; $i=$i+1+$nStep ) {
    $r = ( $r2 - $r1 != 0 ) ? intval( $r1 + ( $r2 - $r1 ) * ( $i / $nRows ) ): $r1;
    $g = ( $g2 - $g1 != 0 ) ? intval( $g1 + ( $g2 - $g1 ) * ( $i / $nRows ) ): $g1;
    $b = ( $b2 - $b1 != 0 ) ? intval( $b1 + ( $b2 - $b1 ) * ( $i / $nRows ) ): $b1;
    if ( "$nOld_r,$nOld_g,$nOld_b" != "$r,$g,$b") {
        $hFill = imagecolorallocate( $hImage, $r, $g, $b );
    }
    imagefilledrectangle($hImage, 0, $i, $nCols, $i+$nStep, $hFill);
    $nOld_r= $r;
    $nOld_g= $g;
    $nOld_b= $b;
}
header("Content-type: image/png");
imagepng($hImage);

Answer 1

我不打算做几何图形 - 而是创建垂直渐变作为更大的图像，然后旋转并裁剪它：

...
$degrees = -5;
$newImage = imagecreatetruecolor($nWidth, $nHeight);
$rotated = imagerotate($hImage, $degrees, 0);
imagecopy($newImage, $rotated, 0, 0, $x, $y, $width, $height)

Answer 2

以下代码片段的运行速度比 GD 库快得多，而且没有复杂性。不过，您必须安装 PHP 的 ImageMagick 扩展。

$oImage = new Imagick();
$oImage->newPseudoImage(1000, 400, 'gradient:#09F-#048' );
$oImage->rotateImage(new ImagickPixel(), -3);
$oImage->cropImage(960, 250, 25, 100);
$oImage->setImageFormat('png');
header( "Content-Type: image/png" );
echo $oImage;