例如a.html页面：

<script>
if(window.ActiveXObject) ajax=new ActiveXObject("Microsoft.XMLHTTP");
else ajax=new XMLHttpRequest();
ajax.open('post',"ajax_link.php?id=1&amp;t="+Math.random(),false);
ajax.send();
document.write("ajax.ResponseText length:"+ajax.responseText.length);
</script>

我怎么能在webkit中抓到此ajax_link.php?id=1&t=xxxxx请求,并获得URL和表单参数？