我想做一个格式如下的HTTP POST,
<?xml version="1.0" encoding="UTF-8" ?>
<authRequest>
<username>someusernamehere</username>
<password>somepasswordhere</password>
</authRequest>
对于任何基于登录的 POST,我通常使用以下机制,
HttpParams params = new BasicHttpParams();
params.setParameter(
"http.useragent",
"Mozilla/5.0 (Windows; U; Windows NT 6.1; en-GB; rv:1.9.2) Gecko/20100115 Firefox/3.6");
DefaultHttpClient httpclient = new DefaultHttpClient(params);
HttpPost httppost = new HttpPost("http://mysite.com/login");
List<NameValuePair> formparams = new ArrayList<NameValuePair>();
formparams.add(new BasicNameValuePair("username", "stackoverflow"));
formparams.add(new BasicNameValuePair("password", "12345"));
UrlEncodedFormEntity entity = new UrlEncodedFormEntity(formparams, "UTF-8");
httppost.setEntity(entity);
HttpResponse httpresponse = httpclient.execute(httppost);
但是通过这种方式,POST 数据看起来像,
username=stackoverflow&password=12345
如何按照我上面提到的指定 XML 格式来格式化这个请求?
提前致谢。