例如: 我怎么能得到“ajax_link.php?id = 1& t = xxxx”这个链接及其网址,使用webkit形成如下?
<script>
if(window.ActiveXObject) ajax=new ActiveXObject("Microsoft.XMLHTTP");
else ajax=new XMLHttpRequest();
ajax.open('post',"ajax_link.php?id=1&t="+Math.random(),false);
ajax.send();
document.write("ajax.ResponseText length:"+ajax.responseText.length);
</script>