ImagickDraw::circle

(PECL imagick 2, PECL imagick 3)

ImagickDraw::circleDraws a circle

说明

public function ImagickDraw::circle(
    float $origin_x,
    float $origin_y,
    float $perimeter_x,
    float $perimeter_y
): bool
警告

本函数还未编写文档,仅有参数列表。

Draws a circle on the image.

参数

origin_x

origin x coordinate

origin_y

origin y coordinate

perimeter_x

perimeter x coordinate

perimeter_y

perimeter y coordinate

返回值

没有返回值。

示例

示例 #1 ImagickDraw::circle() example

<?php
function circle($strokeColor, $fillColor, $backgroundColor, $originX, $originY, $endX, $endY) {

    //Create a ImagickDraw object to draw into.
    $draw = new \ImagickDraw();

    $strokeColor = new \ImagickPixel($strokeColor);
    $fillColor = new \ImagickPixel($fillColor);

    $draw->setStrokeOpacity(1);
    $draw->setStrokeColor($strokeColor);
    $draw->setFillColor($fillColor);

    $draw->setStrokeWidth(2);
    $draw->setFontSize(72);

    $draw->circle($originX, $originY, $endX, $endY);

    $imagick = new \Imagick();
    $imagick->newImage(500, 500, $backgroundColor);
    $imagick->setImageFormat("png");
    $imagick->drawImage($draw);

    header("Content-Type: image/png");
    echo $imagick->getImageBlob();
}

?>

添加备注

用户贡献的备注 1 note

up
11
SkepticaLee
12 years ago
The four values required here are a bit confusing. After all, a circle is defined by three values: the x, y coordinates of the centre, and the radius, r.

The fourth value is redundant, but has to be given, otherwise the function fails. One way of coping with this redundancy is:

<?php
$draw = new ImagickDraw ();
//given that $x and $y are the coordinates of the centre, and $r the radius:
$draw->circle ($x, $y, $x + $r, $y);
?>

There are any number of actions which are synonymous with the last, including:
<?php
$draw->circle ($x, $y, $x, $y + $r);
$draw->circle ($x, $y, $x - $r, $y);
$draw->circle ($x, $y, $x, $y - $r);
// etc, etc.
?>

Hope this helps.
To Top