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.
我正在尝试在我的项目中使用截击来处理所有HTTP请求,因为它是我所知道的最有效的请求。所以我开始学习排球,跟随这个android蜂巢教程。
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 $_POST[""]
as volley sends a JSON object to the URL which we specify.
我的第一个请求是成功的。然后我继续发布请求,但失败了。我在Stack Overflow上看到很多人都有问题,他们结合了post请求和PHP。我认为我们不能使用$_POST["]的常规方式访问它,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.
我尝试了很多方法,但没有成功。我想应该有一种简单而标准的方法来使用PHP。因此,我想知道,为了在PHP代码中接收volley发送的json对象,我需要做什么。
And also how do I check if volley is really sending a JSON object?
如何检查volley是否真的发送JSON对象?
My volley code to send simple post request:
我的截击码发送简单的邮件请求:
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:
<?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');
我尝试了很多在PHP中接受JSON对象的方法,比如这个,以及echo file_get_contents(' PHP://input');
The Result
结果
null
EDIT (The correct way thanks to Georgian Benetatos)
编辑(感谢格鲁吉亚Benetatos的正确方式)
I created the class as you mentioned the class name is CustomRequest
which is as follows:
我创建了类,正如你提到的类名是CustomRequest,如下所示:
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:
在我的活动中,我称之为:
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
5 个解决方案
#1
16
Had a lot of problems myself, try this !
我自己有很多问题,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 $variable Then you build up the response and json_encode it.
你必须做一个映射,映射支持键值类型,而不是你用截击。在php中,你会得到$variable = $_POST["key_from_map"]来retreive它在$变量中的值,然后你构建响应并对其进行json_encode。
Here is a php example of how to query sql and post answer back as JSON
下面是一个php示例,说明如何查询sql并将答案以JSON形式返回
$response["devices"] = array();
while ($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
可以看到,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!
看一下监听器的参数!
#2
6
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:
在服务器端:
<?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.txt
and see the result.
开放的名字。txt并查看结果。
#3
4
Here is a simple code to send post request to php script
下面是发送post请求到php脚本的简单代码。
MainActivity.java
MainActivity.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
volleyRegister.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
源:Android Volley Post请求教程
#4
3
always use StringRequest with volley as it is safer way to get the response from server , if JSON is damaged or not properly formatted.
如果JSON被损坏或格式不正确,那么总是使用volley的StringRequest作为获取服务器响应的更安全的方法。
ANDROID CODE :
ANDROID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> headers = new HashMap<>();
headers.put("name", "Droider");
return headers;
}
@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);
?>
#5
0
This works fine for me if this helps anyone
如果这对任何人都有帮助的话,这对我来说是可行的
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
<?php
$message = $_POST["email"];
echo $message;
?>
#1
16
Had a lot of problems myself, try this !
我自己有很多问题,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 $variable Then you build up the response and json_encode it.
你必须做一个映射,映射支持键值类型,而不是你用截击。在php中,你会得到$variable = $_POST["key_from_map"]来retreive它在$变量中的值,然后你构建响应并对其进行json_encode。
Here is a php example of how to query sql and post answer back as JSON
下面是一个php示例,说明如何查询sql并将答案以JSON形式返回
$response["devices"] = array();
while ($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
可以看到,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!
看一下监听器的参数!
#2
6
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:
在服务器端:
<?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.txt
and see the result.
开放的名字。txt并查看结果。
#3
4
Here is a simple code to send post request to php script
下面是发送post请求到php脚本的简单代码。
MainActivity.java
MainActivity.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
volleyRegister.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
源:Android Volley Post请求教程
#4
3
always use StringRequest with volley as it is safer way to get the response from server , if JSON is damaged or not properly formatted.
如果JSON被损坏或格式不正确,那么总是使用volley的StringRequest作为获取服务器响应的更安全的方法。
ANDROID CODE :
ANDROID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> headers = new HashMap<>();
headers.put("name", "Droider");
return headers;
}
@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);
?>
#5
0
This works fine for me if this helps anyone
如果这对任何人都有帮助的话,这对我来说是可行的
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
<?php
$message = $_POST["email"];
echo $message;
?>