Question

public class MainActivity extends AppCompatActivity {
   String jsonData;
   Response response;
   JSONArray Jarray;
   private String TAG;

   @Override
   protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);
    StrictMode.ThreadPolicy policy = new                                                
    StrictMode.setThreadPolicy(policy);
    OkHttpClient client = new OkHttpClient();
    String token="a";
    RequestBody body = new FormBody.Builder()
            .add("Token",token)
            .build();
    Request request = new Request.Builder()
            .url("http://www.auffers.com/test/index.php")
            .post(body)
            .build();
    Response response = null;
    try {
        response = client.newCall(request).execute();
    } catch (IOException e) {
        e.printStackTrace();
    }
    if (!response.isSuccessful()) try {
        throw new IOException("Unexpected code " + response);
    } catch (IOException e) {
        e.printStackTrace();
    }

    Headers responseHeaders = response.headers();
    for (int i = 0; i < responseHeaders.size(); i++) {
        System.out.println(responseHeaders.name(i) + ": " +      responseHeaders.value(i));
    }

    try {
        System.out.println(response.body().string());

    } catch (IOException e) {
        e.printStackTrace();
    }
}}

给了我{ "name":"John", "age":31, "city":"New York" }但我无法获得不同的元素，如姓名，年龄，城市，以便我可以将它们保存在我的数据库中。我已经研究了许多答案，但却无法做到这一点有人可以帮我解决这个问题吗？

Answer 1

尝试

try {
    JSONObject json = new JSONObject(response.body().string());
    String name = json.getString("name");
    String city = json.getString("city");
    int age = json.getInt("age");

} catch (JSONException e) {
    e.printStackTrace();
}

无法从okhttp中的response.body（）。string（）中获取数据？

1 个答案: