- PHP 手册
- 函数参考
- 图像生成和处理
- ImageMagick
- Imagick
Imagick::rotateImage
(PECL imagick 2, PECL imagick 3)
Imagick::rotateImage — Rotates an image
说明
public Imagick::rotateImage(mixed$background, float $degrees): bool
Rotates an image the specified number of degrees. Empty triangles left over from rotating the image are filled with the background color.
参数
-
background -
The background color
-
degrees -
Rotation angle, in degrees. The rotation angle is interpreted as the number of degrees to rotate the image clockwise.
返回值
成功时返回 true。
更新日志
| 版本 | 说明 |
|---|---|
| PECL imagick 2.1.0 | Now allows a string representing the color as the first parameter. Previous versions allow only an ImagickPixel object. |
范例
示例 #1 Imagick::rotateImage()
<?php
function rotateImage($imagePath, $angle, $color) {
$imagick = new \Imagick(realpath($imagePath));
$imagick->rotateimage($color, $angle);
header("Content-Type: image/jpg");
echo $imagick->getImageBlob();
}
?>
User Contributed Notes 5 notes
up down 6 Anonymous ¶6 years ago
The degrees for imagick and gd is difference!
GD > rotate 90 means counter clockwise.
Imagick > rotate 90 means clockwise.
GD 90 = Imagick 270 or Imagick 90 = GD 270.
Use this function.
<?php
function calculateCounterClockwise($value)
{
if ($value == 0 || $value == 180) {
return $value;
}
if ($value < 0 || $value > 360) {
$value = 90;
}
$total_degree = 360;
$output = intval($total_degree-$value);
return $output;
}// calculateCounterClockwise
echo '1 = '.calculateCounterClockwise(1).'<br>';
echo '90 = '.calculateCounterClockwise(90).'<br>';
echo '270 = '.calculateCounterClockwise(270).'<br>';
echo '359 = '.calculateCounterClockwise(359).'<br>';
echo '360 = '.calculateCounterClockwise(360).'<br>';
?>
test results:
1 = 359
90 = 270
270 = 90
359 = 1
360 = 0
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2
gleb dot deykalo at gmail dot com ¶6 years ago
Some transformations including Imagick ::rotateImage() may change "image page" -- working area inside the image you work on.
Be careful with future modifications afterwards because the image page would be different from new sizes of the image.
For example, if you do Imagic::cropImage() after rotation, you need to set image page properly, otherwise your crop would be performed relating to wrong coordinates (depending on rotation angle, resulting image size may vary).
<?php
$Image = new Imagick($sourceImagePath);
$transparent = '#00000000';
$Image->rotateImage(new \ImagickPixel(), 45); // This makes resulting image bigger
// Set page to be of the full size of new image, starting at top left corner (0, 0)
$Image->setImagePage($Image->getImageWidth(), $Image->getImageHeight(), 0, 0);
$Image->cropImage($crop_w, $crop_h, $crop_x, $crop_y);
?>
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2
AlexG ¶9 years ago
Transparent
<?php $im->rotateImage(new ImagickPixel('#00000000'), 75); ?>
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-1
wjsams at gmail dot com ¶13 years ago
If you want to rotate an image by a certain degree you can do this:
<?php
header('content-type: image/jpeg');
$imagick = new Imagick();
$imagick->readImage('castle.jpg');
$imagick->rotateImage(new ImagickPixel(), 90);
print $imagick->getImage();
?>
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-10
Baptiste VALTHIER ¶11 years ago
You can rotate an jpg image by -13.55° into a transparent png image with :
<?php
$imagick = new Imagick();
$imagick->readImage('my.jpg');
$imagick->rotateImage(new ImagickPixel('none'), -13.55);
$imagick->writeImage('my_rotated.png');
$imagick->clear();
$imagick->destroy();
?>
add a note
官方地址：https://www.php.net/manual/en/imagick.rotateimage.php