HTTP/1.1 302 Moved Temporarily, 这个http是连接超时吗

Question

我模拟登录 提交了一个from表单，header,param 都获取到了，请求的结果怎么是
HTTP/1.1 302 Moved Temporarily 是吗请求的url地址不对么

String url = "http://m44.mm.mail.163.com/auth/login.s?";
service = new RawHttpClientService(HOST,80,false);
ArrayList header = Lists.newArrayList();
ArrayList params = Lists.newArrayList();
header.add(new BasicNameValuePair("Accept-Language", ACCEPT_LANGUAGE));
header.add(new BasicNameValuePair("User-Agent", USER_AGENT));
header.add(new BasicNameValuePair("Accept", "text/html,application/xhtml+xml,application/xml;q=0.9,image/webp,*/*;q=0.8"));
header.add(new BasicNameValuePair("Accept-Charset", "utf-8, iso-8859-1, utf-16, *;q=0.7"));
header.add(new BasicNameValuePair("Accept-Encoding", ACCEPT_ENCODING));
header.add(new BasicNameValuePair("Content-Type", "application/x-www-form-urlencoded"));
header.add(new BasicNameValuePair("Connection:", "keep-alive"));
header.add(new BasicNameValuePair("Origin:", "http://m.mail.163.com"));
header.add(new BasicNameValuePair("Host", HOST));
header.add(new BasicNameValuePair("Cookie:", cookieContainer.toString()));

        params.add(new BasicNameValuePair("method","login"));
        params.add(new BasicNameValuePair("back_url","http://m44.mm.mail.163.com/xm/welcome.do"));
        params.add(new BasicNameValuePair("timestamp","0"));
        params.add(new BasicNameValuePair("m","1"));
        params.add(new BasicNameValuePair("register","0"));
        params.add(new BasicNameValuePair("wml","false"));
        params.add(new BasicNameValuePair("srandid",""));
        params.add(new BasicNameValuePair("ucid",""));
        params.add(new BasicNameValuePair("publicExponent",loginParams.get("publicExponent").toString()));
        params.add(new BasicNameValuePair("modulus",loginParams.get("modulus").toString()));
        params.add(new BasicNameValuePair("username",loginParams.get("username").toString()));
        params.add(new BasicNameValuePair("domain",loginParams.get("domain").toString()));//163.com 126.com yeah.net vip.163.com vip.126.com 188.com
        params.add(new BasicNameValuePair("password",enctry(loginParams.get("publicExponent").toString(),loginParams.get("modulus").toString(),loginParams.get("password").toString())));


        Pair<List<NameValuePair>, String> response = service.request(url,
                HttpMethod.POST,
                header,
                params,
                new DefaultResponseHandle(),
                ENCODE);
        cookieContainer.accept(response.getFirst());

Answer 1

302是跳转，你应该继续请求它返回的跳转地址。