9

为什么我的多个矩形不能在画布中绘制？

<!DOCTYPE html>
<html xmlns="http://www.w3.org/1999/xhtml">
  <head>
    <title></title>
    <script src="Scripts/jquery-1.9.1.min.js"></script>
  </head>
  <body>
    <canvas id="myCanvas" width="500" height="500" style="border:1px solid red">
      <p>Your browser doesn't support canvas.</p>
    </canvas>
  </body>
</html>

<script type ="text/javascript">  
  function Shape(x, y, w, h, fill) {
    this.x = x;
    this.y = y;
    this.w = w;
    this.h = h;
    this.fill = fill;
  }

  // get canvas element.
  var elem = document.getElementById('myCanvas');

  // check if context exist
  if (elem.getContext) {

    var myRect = [];

    myRect.push(new Shape(10, 0, 25, 25, "#333"))
    myRect.push(new Shape(0, 40, 39, 25, "#333"))
    myRect.push(new Shape(0, 80, 100, 25, "#333"))

    context = elem.getContext('2d');
    for (i in myRect) {

      //console.log(x);

      context.fillRect(i.x, i.y, i.w, i.h)
    }
    //// x, y, width, height
    //context.fillRect(0, 0, 50, 50);

    //// x, y, width, height      
    //context.fillRect(75, 0, 50, 50);
  }
</script>

感谢帮助。

4

2 回答 2

16

好的，所以你几乎就在那里。当您遍历矩形数组时，您是在遍历数组键而不是对象本身（请参阅如何以对象为成员循环遍历纯 JavaScript 对象？）。另外，正如@jimjimmy1995 指出的那样，您需要使用.fillStyleas 设置填充。fillRect没有填充参数。

此代码将起作用：

function Shape(x, y, w, h, fill) {
    this.x = x;
    this.y = y;
    this.w = w;
    this.h = h;
    this.fill = fill;
}

// get canvas element.
var elem = document.getElementById('myCanvas');

// check if context exist
if (elem.getContext) {

    var myRect = [];

    myRect.push(new Shape(10, 0, 25, 25, "#333"));
    myRect.push(new Shape(0, 40, 39, 25, "#333"));
    myRect.push(new Shape(0, 80, 100, 25, "#333"));

    context = elem.getContext('2d');
    for (var i in myRect) {
        oRec = myRect[i];
        context.fillStyle = oRec.fill;
        context.fillRect(oRec.x, oRec.y, oRec.w, oRec.h);

    }

}
于 2013-03-05T09:52:10.143 回答
1

你不需要给它填充颜色吗？

context.fillStyle = i.fill;
context.fillRect(i.x,i.y,i.w,i.h);
于 2013-03-05T09:35:55.670 回答