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PHP - Manual: Imagick::rotateImage

2024-12-22

Imagick::rotateImage

(PECL imagick 2, PECL imagick 3)

Imagick::rotateImageRotates 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();
}

?>
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User Contributed Notes 5 notes

up
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
up
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);
?>
up
2
AlexG
9 years ago
Transparent

<?php $im->rotateImage(new ImagickPixel('#00000000'), 75); ?>
up
-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();
?>
up
-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();
?>

官方地址:https://www.php.net/manual/en/imagick.rotateimage.php

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