我的网络服务代码如下我正在使用 WCF Restful 网络服务,
[OperationContract]
[WebInvoke(Method = "POST",
ResponseFormat = WebMessageFormat.Json,
BodyStyle = WebMessageBodyStyle.Wrapped,
UriTemplate = "Login?parameter={parameter}")]
string Login(string parameter);
public string Login(string parameter)
{
/*
* input := {"username":"kevin","password":"123demo"}
* output:= 1=sucess,0=fail
*
*/
//Getting Parameters from Json
JObject jo = JObject.Parse(parameter);
string username = (string)jo["username"];
string password = (string)jo["password"];
return ""+username;
}
我的客户端(Android)代码如下
JSONObject json = new JSONObject();
try {
json.put("username","demo");
json.put("password","password123");
HttpPost postMethod = new HttpPost(SERVICE_URI);
List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();
postMethod.setHeader("Accept", "application/json");
postMethod.setHeader("Content-type", "application/json");
nameValuePairs.add(new BasicNameValuePair("parameter",""+json.toString()));
HttpClient hc = new DefaultHttpClient();
postMethod.setEntity(new UrlEncodedFormEntity(nameValuePairs));
HttpResponse response = hc.execute(postMethod);
Log.i("response", ""+response.toString());
HttpEntity entity = response.getEntity();
final String responseText = EntityUtils.toString(entity);
string=responseText;
Log.i("Output", ""+responseText);
}
catch (Exception e) {
// TODO Auto-generated catch block
Log.i("Exception", ""+e);
}
调用 Web 服务后我得到以下输出:
The server encountered an error processing the request. See server logs for more details.
基本上我的问题是我无法使用 NameValuePair
传递值。
最佳答案
以下代码对我有用:
public static String getJsonData(String webServiceName,String parameter)
{
try
{
String urlFinal=SERVICE_URI+"/"+webServiceName+"?parameter=";
HttpPost postMethod = new HttpPost(urlFinal.trim()+""+URLEncoder.encode(parameter,"UTF-8"));
postMethod.setHeader("Accept", "application/json");
postMethod.setHeader("Content-type", "application/json");
HttpClient hc = new DefaultHttpClient();
HttpResponse response = hc.execute(postMethod);
Log.i("response", ""+response.toString());
HttpEntity entity = response.getEntity();
final String responseText = EntityUtils.toString(entity);
string=responseText;
Log.i("Output", ""+responseText);
}
catch (Exception e) {
}
return string;
}
关于java - 安卓 :-Consume a web service through Post method as a json Request and Response,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/16520666/