Canvas drawImage by points [英] Canvas drawImage by points

问题描述

是否可以使用 canvas.drawImage 但提供dest和source x / y坐标，而不是尺寸？

类似于Java的drawImage的工作原理？

  boolean Graphics.drawImage（Image img，
 int dstx1 ，int dsty1，int dstx2，int dsty2，
 int srcx1，int srcy1，int srcx2，int srcy2，
 ImageObserver observer）; 
  
 
 
解决方案
是的，与其他答案一样：
  drawImage（image，srcX，srcY，srcW，srcH，dstX，dstY，dstW，dstH）; 
  
所以在这种情况下，你必须颠倒目的地和源的顺序， em> width 和 height ：
  boolean Graphics.drawImage b $ b int dstx1，int dsty1，int dstx2，int dsty2，
 int srcx1，int srcy1，int srcx2，int srcy2，
 ImageObserver observer）; 
  
将成为：
  ctx.drawImage（img，srcx1，srcy1，srcx2-srcx1，srcy2  -  srcy1，
 dstx1，dsty1，dstx2-dstx1，dsty2  -  dsty1）; 
  
或作为包装函数，与Java版本兼容参数：
  function drawImage（img，dstx1，dsty1，dstx2，dsty2，srcx1，srcy1，srcx2，srcy2）{
 ctx.drawImage（img，srcx1 ，srcy1，srcx2-srcx1，srcy2-srcy1，
 dstx1，dsty1，dstx2-dstx1，dsty2-dsty1）; 
} 
  
宽度和高度为0无效，源必须在源图像边界。
 
Is it possible to use canvas.drawImage but supplying dest and source x/y co-ordinates, rather than sizes?

Similar to how Java's drawImage works?
boolean Graphics.drawImage(Image img,
   int dstx1, int dsty1, int dstx2, int dsty2,
   int srcx1, int srcy1, int srcx2, int srcy2,
   ImageObserver observer);

解决方案
Yes, as with the other answers:
drawImage(image,   srcX, srcY, srcW, srcH,   dstX, dstY, dstW, dstH);
So in this case you would have to reverse the order of destination and source as well as calculating the width and height:
boolean Graphics.drawImage(Image img,
   int dstx1, int dsty1, int dstx2, int dsty2,
   int srcx1, int srcy1, int srcx2, int srcy2,
   ImageObserver observer);
Will become:
ctx.drawImage(img, srcx1, srcy1, srcx2 - srcx1, srcy2 - srcy1,
                   dstx1, dsty1, dstx2 - dstx1, dsty2 - dsty1);
or as a wrapper function, parameter-compatible with the Java version:
function drawImage(img, dstx1, dsty1, dstx2, dsty2, srcx1, srcy1, srcx2, srcy2) {
    ctx.drawImage(img, srcx1, srcy1, srcx2-srcx1, srcy2-srcy1,
                       dstx1, dsty1, dstx2-dstx1, dsty2-dsty1);
}
Width and height of 0 is invalid, source must be inside the source image boundaries.

                        
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