Android 使用 Volley 发送 post 请求并在 PHP 中接收

Send post request using Volley and receive in PHP

提问by ik024

I am trying to use volley in my project to handle all my HTTP request since it's the most efficient one as far as I know. So I started to learn volley by following this AndroidHive tutorial.

我试图在我的项目中使用 volley 来处理我所有的 HTTP 请求，因为据我所知，它是最有效的。所以我开始按照这个AndroidHive 教程学习 volley 。

My first GET request was successful. Then I moved on to POST request and I failed. I saw on Stack Overflow many people had problems combining post request of volley with PHP. I believe we cannot access it using the normal way that is $_POST[""]as volley sends a JSON object to the URL which we specify.

我的第一个 GET 请求成功了。然后我转到 POST 请求，但失败了。我在 Stack Overflow 上看到很多人在将 volley 的 post 请求与 PHP 结合使用时遇到了问题。我相信我们无法使用正常方式访问它，$_POST[""]因为 volley 将 JSON 对象发送到我们指定的 URL。

There were lots of solutions which I tried but didn't succeed. I guess there should be a simple and standard way of using volley with PHP. So I would like to know what do I need to do in order to receive the json object sent by volley in my PHP code.

有很多我尝试过但没有成功的解决方案。我想应该有一种简单而标准的方式在 PHP 中使用 volley。所以我想知道我需要做什么才能在我的PHP代码中接收由volley发送的json对象。

And also how do I check if volley is really sending a JSON object?

以及如何检查 volley 是否真的在发送 JSON 对象？

My volley code to send simple post request:

我的截击代码发送简单的帖子请求：

JsonObjectRequest jsonObjReq = new JsonObjectRequest(Method.POST,
                url, null,
                new Response.Listener<JSONObject>() {

                    @Override
                    public void onResponse(JSONObject response) {
                        Log.d(TAG, response.toString());
                        pDialog.hide();
                    }
                }, new Response.ErrorListener() {

                    @Override
                    public void onErrorResponse(VolleyError error) {
                        VolleyLog.d(TAG, "Error: " + error.getMessage());
                        pDialog.hide();
                    }
                }) {

            @Override
            protected Map<String, String> getParams() {
                Map<String, String> params = new HashMap<String, String>();
                params.put("name", "Droider");
                return params;
            }

        };

// Adding request to request queue
AppController.getInstance().addToRequestQueue(jsonObjReq, tag_json_obj);

My PHP code to receive json object:(I am pretty sure this is the wrong way, I am not that good in PHP)

我的 PHP 代码接收 json 对象：（我很确定这是错误的方式，我不太擅长 PHP）

<?php
    $jsonReceiveData = json_encode($_POST);
    echo $jsonReceivedData;
?>

I tried lots of ways of accepting JSON object in PHP like this one as well echo file_get_contents('php://input');

我也尝试了很多在 PHP 中接受 JSON 对象的方法，比如这个 echo file_get_contents('php://input');

The Result

结果

null

EDIT (The correct way thanks to Georgian Benetatos)

编辑（正确的方法多亏了Georgian Benetatos）

I created the class as you mentioned the class name is CustomRequestwhich is as follows:

我创建了你提到的类，类名CustomRequest如下：

import java.io.UnsupportedEncodingException;
import java.util.Map;

import org.json.JSONException;
import org.json.JSONObject;

import com.android.volley.NetworkResponse;
import com.android.volley.ParseError;
import com.android.volley.Request;
import com.android.volley.Response;
import com.android.volley.Response.ErrorListener;
import com.android.volley.Response.Listener;
import com.android.volley.toolbox.HttpHeaderParser;

public class CustomRequest extends Request<JSONObject>{

      private Listener<JSONObject> listener;
      private Map<String, String> params;

      public CustomRequest(String url, Map<String, String> params,
                Listener<JSONObject> reponseListener, ErrorListener errorListener) {
            super(Method.GET, url, errorListener);
            this.listener = reponseListener;
            this.params = params;
      }

      public CustomRequest(int method, String url, Map<String, String> params,
                Listener<JSONObject> reponseListener, ErrorListener errorListener) {
            super(method, url, errorListener);
            this.listener = reponseListener;
            this.params = params;
        }

    @Override
    protected Map<String, String> getParams() throws com.android.volley.AuthFailureError {
      return params;
    };

    @Override
    protected void deliverResponse(JSONObject response) {
        listener.onResponse(response);
    }

    @Override
    protected Response<JSONObject> parseNetworkResponse(NetworkResponse response) {
         try {
                String jsonString = new String(response.data,
                        HttpHeaderParser.parseCharset(response.headers));
                return Response.success(new JSONObject(jsonString),
                        HttpHeaderParser.parseCacheHeaders(response));
            } catch (UnsupportedEncodingException e) {
                return Response.error(new ParseError(e));
            } catch (JSONException je) {
                return Response.error(new ParseError(je));
            }
    }

}

Now in my activity I called the following:

现在在我的活动中，我调用了以下内容：

String url = some valid url;
Map<String, String> params = new HashMap<String, String>();
params.put("name", "Droider");

CustomRequest jsObjRequest = new CustomRequest(Method.POST, url, params, new Response.Listener<JSONObject>() {

            @Override
            public void onResponse(JSONObject response) {
                try {
                    Log.d("Response: ", response.toString());
                } catch (JSONException e) {
                    // TODO Auto-generated catch block
                    e.printStackTrace();
                }

            }
        }, new Response.ErrorListener() {

            @Override
            public void onErrorResponse(VolleyError response) {
                Log.d("Response: ", response.toString());
            }
        });
        AppController.getInstance().addToRequestQueue(jsObjRequest);

My PHP code is as follow:

我的PHP代码如下：

<?php
$name = $_POST["name"];

$j = array('name' =>$name);
echo json_encode($j);
?>

Now its returning the correct value:

现在它返回正确的值：

Droider

采纳答案by Georgian Benetatos

Had a lot of problems myself, try this !

自己有很多问题，试试这个！

public class CustomRequest extends Request<JSONObject> {

private Listener<JSONObject> listener;
private Map<String, String> params;

public CustomRequest(String url,Map<String, String> params, Listener<JSONObject> responseListener, ErrorListener errorListener) {
    super(Method.GET, url, errorListener);
    this.listener = responseListener;
    this.params = params;
}

public CustomRequest(int method, String url,Map<String, String> params, Listener<JSONObject> reponseListener, ErrorListener errorListener) {
    super(method, url, errorListener);
    this.listener = reponseListener;
    this.params = params;
}

@Override
protected Map<String, String> getParams() throws com.android.volley.AuthFailureError {
    return params;
};

@Override
protected Response<JSONObject> parseNetworkResponse(NetworkResponse response) {
    try {
        String jsonString = new String(response.data, HttpHeaderParser.parseCharset(response.headers));

        return Response.success(new JSONObject(jsonString), HttpHeaderParser.parseCacheHeaders(response));
    } catch (UnsupportedEncodingException e) {
        return Response.error(new ParseError(e));
    } catch (JSONException je) {
        return Response.error(new ParseError(je));
    }
}

@Override
protected void deliverResponse(JSONObject response) {
    listener.onResponse(response);
}

PHP

PHP

$username = $_POST["username"];
$password = $_POST["password"];

echo json_encode($response);

You have to make a map, the map supports key-value type, and than you post with volley. In php you get $variable = $_POST["key_from_map"] to retreive it's value in the $variable Then you build up the response and json_encode it.

你必须制作一张地图，地图支持键值类型，然后你用凌空发帖。在 php 中你得到 $variable = $_POST["key_from_map"] 来检索它在 $variable 中的值然后你建立响应和 json_encode 它。

Here is a php example of how to query sql and post answer back as JSON

这是一个如何查询 sql 并将答案作为 JSON 发布的 php 示例

$response["devices"] = array();

    while ($row = mysqli_fetch_array($result)) {


        $device["id"] = $row["id"];
        $device["type"] = $row["type"];


        array_push($response["devices"], $device);  
    }

    $response["success"] = true;
    echo json_encode($response);

You can see here that the response type is JSONObject

你可以在这里看到响应类型是 JSONObject

public CustomRequest(int method, String url,Map<String, String> params, Listener<JSONObject> reponseListener, ErrorListener errorListener)

Look at the listener's parameter!

看看监听器的参数！

回答by mmlooloo

JSONObject params = new JSONObject();
        try {
            params.put("name", "Droider");
        } catch (JSONException e) {
            e.printStackTrace();
        }
JsonObjectRequest jsonObjReq = new JsonObjectRequest(Method.POST,
                url, params,
                new Response.Listener<JSONObject>() {

                    @Override
                    public void onResponse(JSONObject response) {
                        Log.d(TAG, response.toString());
                        pDialog.hide();
                    }
                }, new Response.ErrorListener() {

                    @Override
                    public void onErrorResponse(VolleyError error) {
                        VolleyLog.d(TAG, "Error: " + error.getMessage());
                        pDialog.hide();
                    }
                }) {

                 @Override
                 public Map<String, String> getHeaders() throws AuthFailureError {
                        HashMap<String, String> headers = new HashMap<String, String>();
                        headers.put("Content-Type", "application/json; charset=utf-8");
                        return headers;  
                 } 

        };

// Adding request to request queue
AppController.getInstance().addToRequestQueue(jsonObjReq, tag_json_obj);

and in your server side:

并在您的服务器端：

<?php
     $value = json_decode(file_get_contents('php://input'));
     $file = 'MyName.txt';
     file_put_contents($file, "The received name is {$value->name} ", FILE_APPEND | LOCK_EX);    
?>

open MyName.txtand see the result.

打开MyName.txt看看结果。

回答by Belal Khan

Here is a simple code to send post request to php script

这是一个向php脚本发送post请求的简单代码

MainActivity.java

主活动.java

public class MainActivity extends AppCompatActivity implements View.OnClickListener {

private static final String REGISTER_URL = "http://simplifiedcoding.16mb.com/UserRegistration/volleyRegister.php";

public static final String KEY_USERNAME = "username";
public static final String KEY_PASSWORD = "password";
public static final String KEY_EMAIL = "email";


private EditText editTextUsername;
private EditText editTextEmail;
private EditText editTextPassword;

private Button buttonRegister;

@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);

    editTextUsername = (EditText) findViewById(R.id.editTextUsername);
    editTextPassword = (EditText) findViewById(R.id.editTextPassword);
    editTextEmail= (EditText) findViewById(R.id.editTextEmail);

    buttonRegister = (Button) findViewById(R.id.buttonRegister);

    buttonRegister.setOnClickListener(this);
}

private void registerUser(){
    final String username = editTextUsername.getText().toString().trim();
    final String password = editTextPassword.getText().toString().trim();
    final String email = editTextEmail.getText().toString().trim();

    StringRequest stringRequest = new StringRequest(Request.Method.POST, REGISTER_URL,
            new Response.Listener<String>() {
                @Override
                public void onResponse(String response) {
                    Toast.makeText(MainActivity.this,response,Toast.LENGTH_LONG).show();
                }
            },
            new Response.ErrorListener() {
                @Override
                public void onErrorResponse(VolleyError error) {
                    Toast.makeText(MainActivity.this,error.toString(),Toast.LENGTH_LONG).show();
                }
            }){
        @Override
        protected Map<String,String> getParams(){
            Map<String,String> params = new HashMap<String, String>();
            params.put(KEY_USERNAME,username);
            params.put(KEY_PASSWORD,password);
            params.put(KEY_EMAIL, email);
            return params;
        }

    };

    RequestQueue requestQueue = Volley.newRequestQueue(this);
    requestQueue.add(stringRequest);
}

@Override
public void onClick(View v) {
    if(v == buttonRegister){
        registerUser();
    }
}
}

volleyRegister.php

排球注册.php

<?php

if($_SERVER['REQUEST_METHOD']=='POST'){
    $username = $_POST['username'];
    $email = $_POST['email'];
    $password = $_POST['password'];

    require_once('dbConnect.php');

    $sql = "INSERT INTO volley (username,password,email) VALUES ('$username','$email','$password')";


    if(mysqli_query($con,$sql)){
        echo "Successfully Registered";
    }else{
        echo "Could not register";

    }
}else{
echo 'error'}
}

Source: Android Volley Post Request Tutorial

来源：Android Volley Post 请求教程

回答by Raghav Satyadev

always use StringRequest with volley as it is safer way to get the response from server , if JSON is damaged or not properly formatted.

始终将 StringRequest 与 volley 一起使用，因为如果 JSON 损坏或格式不正确，这是从服务器获取响应的更安全的方法。

ANDROID CODE :

安卓代码：

StringRequest stringRequest = new StringRequest(Request.Method.POST, URL, new Response.Listener<String>() {
        @Override
        public void onResponse(String response) {
            try {JSONObject jsonObject = new JSONObject(response);
            } catch (JSONException ignored) {
            }
        }
    }, new Response.ErrorListener() {
        @Override
        public void onErrorResponse(VolleyError volleyError) {
            if (volleyError instanceof TimeoutError) {
            }
        }
    }) {
        @Override
        public Map<String, String> getParams() throws AuthFailureError {
            HashMap<String, String> params = new HashMap<>();
            params.put("name", "Droider");
            return params;
        }

        @Override
        public Priority getPriority() {
            return Priority.IMMEDIATE;
        }
    };
    ApplicationController.getInstance().addToRequestQueue(stringRequest);

PHP CODE :

PHP代码：

<?php
   $name = $_POST["name"];
   $j = array('name' =>$name);
   echo json_encode($j);
?>

回答by sha1962

This works fine for me if this helps anyone

如果这对任何人有帮助，这对我来说很好用

public class LoginActivity extends AppCompatActivity {

    private EditText Email;
    private EditText Password;
    private String URL = "http://REPLACE ME WITH YOUR URL/login.php";

    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_login);

        TextView register = (TextView) findViewById(R.id.Register);
        TextView forgotten = (TextView) findViewById(R.id.Forgotten);
        Button login = (Button) findViewById(R.id.Login);
        Email = (EditText) findViewById(R.id.Email);
        Password = (EditText) findViewById(R.id.Password);
        Password.setImeOptions(EditorInfo.IME_ACTION_DONE);

        login.setOnClickListener(new View.OnClickListener() {
            @Override
            public void onClick(View v) {
                RequestQueue MyRequestQueue = Volley.newRequestQueue  (LoginActivity.this);
                MyRequestQueue.add(MyStringRequest);
            }
        });
    }

StringRequest MyStringRequest = new StringRequest(Request.Method.POST, URL, new Response.Listener<String>() {

        @Override
        public void onResponse(String response) {
            Toast.makeText(getApplicationContext(),response.trim(), Toast.LENGTH_SHORT).show();
        }

    }, new Response.ErrorListener() { //Create an error listener to handle errors appropriately.
        @Override
        public void onErrorResponse(VolleyError error) {
            Toast.makeText(getApplicationContext(),error.toString().trim(), Toast.LENGTH_LONG).show();
        }
    }) {
        protected Map<String, String> getParams() {
            final String email = Email.getText().toString().trim();
            final String password = Password.getText().toString().trim();

            Map<String, String> MyData = new HashMap<String, String>();
            MyData.put("email", email);
            MyData.put("password", password);
            return MyData;
        }
     };
 }

the login.php

登录.php

<?php
    $message = $_POST["email"];
    echo $message;
?>