在Android上，我正在尝试使用简单的JSON对象请求一个简单的POST-API。
说，我有一个简单的API（1.1.1.1/testApi），它响应了一个包含以下内容的JSON对象：
Calling the API using Postman就像一个魅力，所以我认为我的API很好。
我已经尝试了下面的一些链接：
CallApi对象并解析API地址（例如URL）的示例，因此当我尝试调用对象时总会出错。我确实花时间搜索有关此问题的解决方案，但是没有“简单”方法来调用POST-API。
是否有任何简单的方法接受URL输入，然后返回JSON对象？
如果这是一个重复的问题，请告诉我。
提前致谢。
答案 0 :(得分：2)
您好我有工作改造示例以您的方式尝试
让我们开始
1） Gradle
compile 'com.google.code.gson:gson:2.2.4'
compile 'com.squareup.okhttp:okhttp:2.4.0'
compile 'com.squareup.retrofit:retrofit:2.0.0-beta2'
compile 'com.squareup.retrofit:converter-gson:2.0.0-beta2'
compile 'com.squareup.okhttp3:logging-interceptor:3.0.1'
2）界面
public interface ServiceInterface {
@GET(HttpConstants.USERDATAJSON)
Call<ListData>taskData(@Query("method")String method,@Query("stdID")int stdID);
}
3）服务类
public class ServiceClass {
static ServiceInterface serviceInterface;
// public static final String baseUrl= HttpConstants.BASE_URL_GEONAME;
public static final String baseUrl= HttpConstants.baseUrl;
public static ServiceInterface connection(){
if(serviceInterface==null){
HttpLoggingInterceptor interceptor = new HttpLoggingInterceptor();
interceptor.setLevel(HttpLoggingInterceptor.Level.BODY);
OkHttpClient client = new OkHttpClient();
client.interceptors().add(new Interceptor() {
@Override
public Response intercept(Chain chain) throws IOException {
Response response=chain.proceed(chain.request());
return response;
}
});
Retrofit retrofit = new Retrofit.Builder()
.client(client)
.addConverterFactory(GsonConverterFactory.create())
.baseUrl(baseUrl)
.build();
serviceInterface=retrofit.create(ServiceInterface.class);
}
return serviceInterface;
}
}
4）从活动中调用方法
public void getTaskData(){
ServiceInterface serviceInterface=ServiceClass.connection();
Call<ListData> call=serviceInterface.taskData("getAllUsersSimple",0);
call.enqueue(new Callback<ListData>() {
@Override
public void onResponse(Response<ListData> response, Retrofit retrofit) {
Log.v("@@@Response",""+response.toString());
if(response.isSuccess()){
listData=response.body();
dataList=listData.getData();
printStudentDetails(dataList);
}
}
@Override
public void onFailure(Throwable t) {
Log.v("@@@Failure"," Message"+t.getMessage());
}
});
}
5） Pojo
public class ListData {
@SerializedName("data")
@Expose
private List<DataPojo> data = null;
@SerializedName("code")
@Expose
private Integer code;
@SerializedName("message")
@Expose
private String message;
public List<DataPojo> getData() {
return data;
}
public void setData(List<DataPojo> data) {
this.data = data;
}
public Integer getCode() {
return code;
}
public void setCode(Integer code) {
this.code = code;
}
public String getMessage() {
return message;
}
public void setMessage(String message) {
this.message = message;
}
}
public class DataPojo {
@SerializedName("user_id")
@Expose
private String userId;
@SerializedName("user_name")
@Expose
private String userName;
@SerializedName("user_age")
@Expose
private String userAge;
public String getUserId() {
return userId;
}
public void setUserId(String userId) {
this.userId = userId;
}
public String getUserName() {
return userName;
}
public void setUserName(String userName) {
this.userName = userName;
}
public String getUserAge() {
return userAge;
}
public void setUserAge(String userAge) {
this.userAge = userAge;
}
}
您可以使用此链接创建您的pojo http://www.jsonschema2pojo.org/
有关更多参考，请访问该链接 https://github.com/pratikvyas1991/NetworkingExample/tree/master/app
答案 1 :(得分：1)
拉玛特。如果您希望向Web API发送POST请求，可以尝试使用Android Volley Library。您可以参考以下链接。
Android排球图书馆
教程
答案 2 :(得分：1)
您可以使用 Restful 服务使用 RestTemplate ，这非常简单。下面是一个示例代码，我在其中发布了一个Object。
public MasterObject setMasterByBatch(MasterObject masterObject) {
try {
RestTemplate restTemplate = new RestTemplate();
restTemplate.getMessageConverters().add(new MappingJackson2HttpMessageConverter());
StrictMode.ThreadPolicy policy = new StrictMode.ThreadPolicy.Builder().permitAll().build();
StrictMode.setThreadPolicy(policy);
masterObject = restTemplate.postForObject(yourUrl, masterObject, MasterObject.class);
} catch (Exception e) {
e.printStackTrace();
Log.e("masterObjPost_WsCli_EX", e.toString());
}
return masterObject;
}
在build.gradle（Module：app）中需要很少的依赖项：
dependencies {
compile 'org.springframework.android:spring-android-rest-template:1.0.1.RELEASE'
compile 'com.fasterxml.jackson.core:jackson-core:2.6.0'
compile 'com.fasterxml.jackson.core:jackson-databind:2.6.0'
}
如果显示有关 org.springframework 的任何错误，您可能需要下载并插入弹簧库
答案 3 :(得分：1)
AsyncTask示例
我个人也更喜欢Retrofit / Volley，具体取决于项目需求。
如果要为您设置标题（testApi）Rest API。（基本授权）
String credentials = email + ":" + password;
String basicAuth = "Basic " + new String(new Base64().encode(credentials.getBytes()));
connection.setRequestProperty ("Authorization", basicAuth);
connection..setRequestProperty("Content-Language", "en-US");
注意：
无法在主线程中进行网络操作/调用。您需要从另一个线程，异步任务或意向服务
所有UI操作都应该在onPostExecute，onPreExecute
在您想要的地方调用AsyncTask
以下代码可能会对您有所帮助。
import android.app.Activity;
import android.os.AsyncTask;
import android.support.v7.app.AppCompatActivity;
import android.os.Bundle;
import android.widget.Toast;
import org.json.JSONObject;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.net.HttpURLConnection;
import java.net.URL;
import java.net.URLEncoder;
import java.util.HashMap;
import java.util.Map;
public class MainActivity extends AppCompatActivity {
String TEST_URL="http://172.16.68.4:8080/testApi";
Activity activity;
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
activity=MainActivity.this;
new PostAsyncTask().execute();
}
private class PostAsyncTask extends AsyncTask<String,Void,JSONObject> {
@Override
protected void onPreExecute() {
super.onPreExecute();
}
@Override
protected JSONObject doInBackground(String... params) {
String value="test";
Map postData = new HashMap<>();
postData.put("key",value);
return post(TEST_URL,postData);
}
@Override
protected void onPostExecute(JSONObject response) {
super.onPostExecute(response);
//All your UI operation can be performed here
//Response string can be converted to JSONObject/JSONArray like
try {
Toast.makeText(activity, String.format("%s : %s",response.getString("status"),response.getString("name")), Toast.LENGTH_LONG).show();
} catch (Exception e) {
e.printStackTrace();
Toast.makeText(activity, String.format("%s","Something went wrong!!!!!!"), Toast.LENGTH_LONG).show();
}
System.out.println(response);
}
}
/**
* Method allows to HTTP POST request to the server to send data to a specified resource
* @param REQUEST_URL URL of the API to be requested
* @param params parameter that are to be send in the "body" of the request Ex: parameter=value&also=another
* returns response as a JSON object
*/
public JSONObject post(String REQUEST_URL,Map<String, Object> params) {
JSONObject jsonObject = null;
BufferedReader reader = null;
try { URL url = new URL(REQUEST_URL);
StringBuilder postData = new StringBuilder();
for (Map.Entry<String, Object> param : params.entrySet()) {
if (postData.length() != 0) postData.append('&');
postData.append(URLEncoder.encode(param.getKey(), "UTF-8"));
postData.append('=');
postData.append(URLEncoder.encode(String.valueOf(param.getValue()), "UTF-8"));
}
byte[] postDataBytes = postData.toString().getBytes("UTF-8");
HttpURLConnection connection = (HttpURLConnection) url.openConnection();
connection.setRequestProperty("Content-Type", "application/x-www-form-urlencoded");
connection.setConnectTimeout(8000);
connection.setRequestMethod("POST");
connection.setUseCaches(false);
connection.setDoOutput(true);
connection.getOutputStream().write(postDataBytes);
connection.connect();
StringBuilder sb;
int statusCode = connection.getResponseCode();
if (statusCode == 200) {
sb = new StringBuilder();
reader = new BufferedReader(new InputStreamReader(connection.getInputStream()));
String line;
while ((line = reader.readLine()) != null) {
sb.append(line);
}
jsonObject = new JSONObject(sb.toString());
}
connection.disconnect();
} catch (Exception e) {
e.printStackTrace();
} finally {
if (reader != null) {
try {
reader.close();
} catch (Exception e) {
e.printStackTrace();
}
}
}
return jsonObject;
}
}
答案 4 :(得分：0)
就个人而言，我更喜欢Retrofit，使用
非常简单，非常好用