xmlhttprequest状态302的问题

Question

我试图在链接megaupload中获得真正的路径,但总是但这不起作用.

function getRealURL(){

    var st = new String(""); 
    var req = new XMLHttpRequest();
    req.open("GET","http://www.megaupload.com/?d=6CKP1MVJ",true);
    req.send(null);
    req.send(null);
    req.onreadystatechange = function (aEvt) {
     if (req.readyState == 4) {
        if(req.status == 302){
          //SUCESSO
           st = req.responseText;
        }
      }
    };//funcao

    element.getElementById("id").setAttribute("value", st);

}

我需要这个链接:重定向到:http://www534.megaupload.com/files/c2c36829bc392692525f5b7b3d9d81dd/Coldplay - 警告Sign.mp3

这是:http://www.megaupload.com/？ d = 6CKP1MVJ

Answer 1

XMLHttpRequest默认情况下会自动遵循重定向,因此您不会看到302响应.您需要将nsIHttpChannel.redirectionLimit属性设置为零以防止它:

req.open("GET","http://www.megaupload.com/?d=6CKP1MVJ",true);
req.channel.QueryInterface(Components.interfaces.nsIHttpChannel).redirectionLimit = 0;
req.send(null);

并不是说你在这里使用的链接会重定向到任何地方,但这是一般方法.顺便说一下,你应该看看重定向的响应文本,而不是查看req.getResponseHeader("Location").