如何发送和接收Android和ASP.Net MVC的JSON数据？ 我有一个Android登录应用程序，它有一个Login活动类和一个JSONParser类。 使用JSONParser类，我传递了mvc位置的URL，＆＃34; POST＆＃34;参数和我的参数用户名，密码为json到MVC。 我有一个接受用户名和密码的ASP.net mvc代码，如果找到匹配，它将返回json数据为＆＃34; username＆＃34; ：＆＃34;管理员，＆＃34;成功＆＃34; ：1。
Login Activity类的代码是：
protected String doInBackground(String... arg0) {
// TODO Auto-generated method stub
new Thread() {
// Running Thread.
public void run() {
List<NameValuePair> params = new ArrayList<NameValuePair>();
params.add(new BasicNameValuePair("userid", username.getText().toString().trim()));
params.add(new BasicNameValuePair("password", password.getText().toString().trim()));
JSONObject json = jsonParser.makeHttpRequest("http://localhost:8012/Login/Login","GET", params);
Log.d("Create Response", json.toString());
try {
int success = json.getInt("success");
if (success == 1) {
Intent newregistrationIntent = new Intent(MainActivity.this,mydashActivity.class);
startActivityForResult(newregistrationIntent, 0);
}
else
{
i=1;
flag=1;
}
} catch (JSONException e) {
e.printStackTrace();
}
}
}.start();
return null;
}
JSONParser的代码是：
public class JSONParser {
static InputStream is = null;
static JSONObject jObj = null;
static String json = "";
// constructor
public JSONParser() {
}
// function get json from url
// by making HTTP POST or GET mehtod
public JSONObject makeHttpRequest(String url, String method,
List<NameValuePair> params) {
// Making HTTP request
try {
// check for request method
if(method == "POST"){
// request method is POST
// defaultHttpClient
DefaultHttpClient httpClient = new DefaultHttpClient();
HttpPost httpPost = new HttpPost(url);
httpPost.setEntity(new UrlEncodedFormEntity(params));
httpPost.setHeader("Content-type", "application/json");
HttpResponse httpResponse = httpClient.execute(httpPost);
HttpEntity httpEntity = httpResponse.getEntity();
is = httpEntity.getContent();
}else if(method == "GET"){
// request method is GET
DefaultHttpClient httpClient = new DefaultHttpClient();
String paramString = URLEncodedUtils.format(params, "utf-8");
url += "?" + paramString;
HttpGet httpGet = new HttpGet(url);
HttpResponse httpResponse = httpClient.execute(httpGet);
HttpEntity httpEntity = httpResponse.getEntity();
is = httpEntity.getContent();
}
} catch (UnsupportedEncodingException e) {
e.printStackTrace();
} catch (ClientProtocolException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
try {
BufferedReader reader = new BufferedReader(new InputStreamReader(
is, "iso-8859-1"), 8);
StringBuilder sb = new StringBuilder();
String line = null;
while ((line = reader.readLine()) != null) {
sb.append(line + "\n");
}
is.close();
json = sb.toString();
} catch (Exception e) {
Log.e("Buffer Error", "Error converting result " + e.toString());
}
// try parse the string to a JSON object
try {
jObj = new JSONObject(json);
} catch (JSONException e) {
Log.e("JSON Parser", "Error parsing data " + e.toString());
}
// return JSON String
return jObj;
}
}
我的ASP.Net MVC代码是：
using System;
using System.Collections.Generic;
using System.Linq;
using System.Web;
using System.Web.Mvc;
using MvcApplication1.Models;
using System.Web.Script.Serialization;
namespace MvcApplication1.Controllers
{
public class LoginController : Controller
{
Dictionary<string, object> loginParam = new Dictionary<string, object>();
//
// GET: /Login/
[HttpGet]
public ActionResult Login()
{
return View();
}
[HttpPost]
public ActionResult Login(Login cs)
{
return Json(new { username="admin", success = 1 });
}
[HttpGet]
public ActionResult Profile()
{
return View();
}
[HttpPost]
public ActionResult Profile(Login cs)
{
return View("");
}
}
}
我不确定是否实际上是从android到mvc代码的连接。我也不确定MVC控制器是否被安卓代码命中了Login（）。 如何确保JSON数据是从Android发送的，以及从MVC代码返回的数据？
注意：我实际上并不是在比较MVC代码中的数据。我只是返回数据。这是我的第一个MVC代码。
答案 0 :(得分：0)
public static String POST( String username,String password){
String url=Constants.url_registration;
InputStream inputStream = null;
String result = "";
try {
// 1. create HttpClient
HttpClient httpclient = new DefaultHttpClient();
// 2. make POST request to the given URL
HttpPost httpPost = new HttpPost(url);
String json = "";
// 3. build jsonObject
JSONObject jsonObject = new JSONObject();
jsonObject.accumulate("username", username);
jsonObject.accumulate("password", password);
// 4. convert JSONObject to JSON to String
json = jsonObject.toString();
// ** Alternative way to convert Person object to JSON string usin Jackson Lib
// ObjectMapper mapper = new ObjectMapper();
// json = mapper.writeValueAsString(person);
// 5. set json to StringEntity
StringEntity se = new StringEntity(json);
// 6. set httpPost Entity
httpPost.setEntity(se);
// 7. Set some headers to inform server about the type of the content
httpPost.setHeader("Accept", "application/json");
httpPost.setHeader("Content-type", "application/json");
// 8. Execute POST request to the given URL
HttpResponse httpResponse = httpclient.execute(httpPost);
// 9. receive response as inputStream
inputStream = httpResponse.getEntity().getContent();
// 10. convert inputstream to string
if(inputStream != null)
result = convertInputStreamToString(inputStream);
else
result = "Did not work!";
} catch (Exception e) {
Log.d("InputStream", e.getLocalizedMessage());
}
// 11. return result
return result;
}
答案 1 :(得分：0)
因为您使用了“获取”方法：
JSONObject json = jsonParser.makeHttpRequest("http://localhost:8012/Login/Login","GET", params);
虽然登录方法是[httpPost]
[HttpPost]
public ActionResult Login(Login cs)
{
return Json(new { username="admin", success = 1 });
}