使用js

时间:2018-04-25 18:46:00

标签: javascript html

我有一个网页,我需要一个按钮,其中截取屏幕截图。现在,我有以下js函数,它被html上的按钮驱动:

var takeScreenShot = function() {
    html2canvas(document.getElementById("container"), {
        onrendered: function (canvas) {
            var tempcanvas=document.createElement('canvas');
            tempcanvas.width=550;
            tempcanvas.height=650;
            var context=tempcanvas.getContext('2d');
            context.drawImage(canvas,0,0,tempcanvas.width,tempcanvas.height,0,0,tempcanvas.width,tempcanvas.height);
            var link=document.createElement("a");
            link.href=tempcanvas.toDataURL('image/jpg');
            link.download = 'screenshot.jpg';
            link.click();
        }
    });
}

然后,html具有以下内容:

<div style="margin-right: 100px; margin-left: 600px" id="container">
  <table class="myTable">
    <th class="titles">Title</th>
    <!-- code code code -->
    <button onclick="takeScreenShot()">Table Screenshot</button>
  </table>
</div>

这样可行,但问题是所拍摄的屏幕出现错误。下一张图片是如何截取屏幕截图:

Screenshot taken with the button

这就是它应该如何。

Screenshot taken manually

0 个答案:

没有答案