Din*_*esh 4 java http-post apache-commons-httpclient
我想做格式的HTTP POST,如下所示,
<?xml version="1.0" encoding="UTF-8" ?>
<authRequest>
<username>someusernamehere</username>
<password>somepasswordhere</password>
</authRequest>
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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);
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但是通过这种方式,POST数据看起来像,
username=stackoverflow&password=12345
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如何根据我上面提到的指定XML格式格式化此请求?
提前致谢.
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