JCropped获取裁剪的图像高度 [英] JCropped Get Cropped Image Height

问题描述

我正在使用JCrop来获取要裁剪的图像的尺寸.我有用于获取尺寸集的代码，如下所示:

I'm using JCrop to get the dimensions of the image to be cropped. I have the code for getting the dimensions set like this:

            function updateCoordinates(c) {
                console.log(c)
                $('#image_x').val(c.x);
                $('#image_x2').val(c.x2);
                $('#image_y').val(c.y);
                $('#image_y2').val(c.y2);
                $('#image_w').val(c.w);
            }


            $('#jcrop_target').Jcrop({
                aspectRatio : 2 / 1,
                onSelect : updateCoordinates,
                onChange : updateCoordinates
            });

变量'c'具有值c.x，c.x2，c.y，c.y2和c.w.我正在猜测c.w =宽度，c.x = x坐标，c.y = y坐标，但是如何获得图像的高度?我很想将这些值传递给PHP Imagick裁剪.

the variable 'c' has the values c.x, c.x2, c.y, c.y2 and c.w . I'm making a guess that c.w = width, c.x = x coordinates and c.y = y coordinates, but how do I get the height of the image? I am cry to pass the values into PHP Imagick cropping.

<?php    
$picture = new Imagick($this -> getOriginalUrl());
$picture->cropImage($width, $height, $x, $y);

我觉得我缺少明显的东西.

I Feel like I'm missing something obvious.

推荐答案

c.y2-c.y给出裁剪区域的高度.

c.y2 - c.y gives the height of the cropped area.

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