open动态修改img的onclick事件示例代码_javascript技巧
时间:2021-07-01 10:21:17
帮助过:5人阅读
代码如下:
var imgsrc = document.getElementById("imgsrc").getElementsByTagName('img');
imgsrc[0].style.cursor="hand";
imgsrc[0].onclick = new Function( "openArticle('/RssCommServlet?catalogid=29')");
imgsrc[1].style.cursor="hand";
imgsrc[1].onclick = new Function( "openArticle('/RssCommServlet?catalogid=30')");
imgsrc[2].style.cursor="hand";
imgsrc[2].onclick = new Function( "openArticle('/RssCommServlet?catalogid=31')");