使用post方法volley将jsonobject作为请求参数传递 [英] Pass jsonobject as request parameter using post method volley
本文介绍了使用post方法volley将jsonobject作为请求参数传递的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
问题描述
我只是试图使用JsonRequestObject将值发送到php脚本并接收json数据，但是下面的代码不起作用
I was just trying to use JsonRequestObject to send values to php script and receive json data but below code is not working
package com.demo.volleyjsondemo;
import android.content.Context;
import android.support.v7.app.AppCompatActivity;
import android.os.Bundle;
import android.util.Log;
import android.widget.TextView;
import com.android.volley.AuthFailureError;
import com.android.volley.Request;
import com.android.volley.Response;
import com.android.volley.VolleyError;
import com.android.volley.toolbox.JsonObjectRequest;
import com.android.volley.toolbox.StringRequest;
import com.demo.volleyjsondemo.Utils.Constants;
import com.demo.volleyjsondemo.Utils.RequestSingleTone;
import org.json.JSONException;
import org.json.JSONObject;
import java.util.HashMap;
import java.util.Map;
public class MainActivity extends AppCompatActivity {
JsonObjectRequest jsonObjectRequest;
TextView txtName, txtEmail;
JSONObject parameters;
Context context;
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
init();
try {
sendReq();
} catch (JSONException e) {
e.printStackTrace();
}
}
private void init() {
context = MainActivity.this;
txtName = (TextView) findViewById(R.id.txtName);
txtEmail = (TextView) findViewById(R.id.txtEmail);
}
private void sendReq() throws JSONException {
//just to demonstrate how to send parameters with json request
parameters = new JSONObject();
try {
parameters.put(Constants.NAME, "jack");
Log.e("paramter",parameters.toString());
} catch (JSONException e) {
e.printStackTrace();
}
//jsonobjectrequest to send request and get response in json
jsonObjectRequest = new JsonObjectRequest(
Request.Method.POST //request method
, Constants.BASE_URL + Constants.GET_PERSON_DETAILS_URL //URL of php file
, new JSONObject(parameters.toString()) //parameters to send to server
, new Response.Listener<JSONObject>() { //response will come here in case of success
@Override
public void onResponse(JSONObject response) {
try {
Log.e("response", response.getString(Constants.NAME));
} catch (JSONException e) {
e.printStackTrace();
}
}
},
new Response.ErrorListener() { //response will come here in case of error
@Override
public void onErrorResponse(VolleyError error) {
error.printStackTrace();
}
});
//add request to singletone
RequestSingleTone.getInstance(context).addRequest(jsonObjectRequest);
}
}
php代码
<?php
require_once 'dbconfig.php';
$name = $_POST['name'];
//$name="jack";
$singlePersonInfoQuery = "SELECT * FROM test WHERE name='".$name."'";
$result = mysqli_query($con,$singlePersonInfoQuery);
if(mysqli_num_rows($result) > 0){
$raw = mysqli_fetch_assoc($result);
echo json_encode(array("name"=>$raw['name'],"email"=>$raw['email']));
}else{
echo json_encode(array("name"=>$name,"email"=>"blank"));
}
?>
我遇到以下错误
10-06 14:36:34.943 5256-5256/com.demo.volleyjsondemo W/System.err: com.android.volley.ParseError: org.json.JSONException: Value perfect<br of type java.lang.String cannot be converted to JSONObject
10-06 14:36:34.943 5256-5256/com.demo.volleyjsondemo W/System.err: at com.android.volley.toolbox.JsonObjectRequest.parseNetworkResponse(JsonObjectRequest.java:73)
10-06 14:36:34.943 5256-5256/com.demo.volleyjsondemo W/System.err: at com.android.volley.NetworkDispatcher.run(NetworkDispatcher.java:123)
10-06 14:36:34.943 5256-5256/com.demo.volleyjsondemo W/System.err: Caused by: org.json.JSONException: Value perfect<br of type java.lang.String cannot be converted to JSONObject
10-06 14:36:34.943 5256-5256/com.demo.volleyjsondemo W/System.err: at org.json.JSON.typeMismatch(JSON.java:111)
10-06 14:36:34.943 5256-5256/com.demo.volleyjsondemo W/System.err: at org.json.JSONObject.<init>(JSONObject.java:160)
10-06 14:36:34.943 5256-5256/com.demo.volleyjsondemo W/System.err: at org.json.JSONObject.<init>(JSONObject.java:173)
10-06 14:36:34.943 5256-5256/com.demo.volleyjsondemo W/System.err: at com.android.volley.toolbox.JsonObjectRequest.parseNetworkResponse(JsonObjectRequest.java:68)
10-06 14:36:34.943 5256-5256/com.demo.volleyjsondemo W/System.err: ... 1 more
我猜上面的警告是因为在php脚本中接收参数的方式错误.如果我不传递任何参数或使用stringrequest但我想使用jsonrequestobject，以上代码可以正常工作.是错误的
I guess about above warning is because of wrong way of receiving parameters in php script..Above code working fine if i don't pass any parameters or using stringrequest but i want to use jsonrequestobject..I can't understand what is wrong
推荐答案
最后回答了我自己的问题，调试了我的php脚本后，我发现我已经完成了在php脚本中接收参数的操作.$ _POST或$ _GET可以使用.请在下面参考，该方法可以正常使用..我希望它可以对某人有所帮助
Finally answering my own question,After debugging my php script i found that i done in receiving parameters in php script..As i am sending parameters as jsonobject,No $_POST or $_GET will work.Please refer below which is working fine..I hope it will help someone
android代码:-
android code:-
package com.demo.volleyjsondemo;
import android.content.Context;
import android.support.v7.app.AppCompatActivity;
import android.os.Bundle;
import android.util.Log;
import android.widget.TextView;
import com.android.volley.AuthFailureError;
import com.android.volley.Request;
import com.android.volley.Response;
import com.android.volley.VolleyError;
import com.android.volley.toolbox.JsonObjectRequest;
import com.android.volley.toolbox.StringRequest;
import com.demo.volleyjsondemo.Utils.Constants;
import com.demo.volleyjsondemo.Utils.RequestSingleTone;
import org.json.JSONException;
import org.json.JSONObject;
import java.io.UnsupportedEncodingException;
import java.util.HashMap;
import java.util.Map;
public class MainActivity extends AppCompatActivity {
JsonObjectRequest jsonObjectRequest;
TextView txtName, txtEmail;
Context context;
Map<String, String> parametersMap = new HashMap<>();
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
init();
try {
sendReq();
} catch (JSONException e) {
e.printStackTrace();
}
}
private void init() {
context = MainActivity.this;
txtName = (TextView) findViewById(R.id.txtName);
txtEmail = (TextView) findViewById(R.id.txtEmail);
}
private void sendReq() throws JSONException {
//just to demonstrate how to send parameters with json request
parametersMap.put(Constants.NAME, "jack");
//jsonobjectrequest to send request and get response in json
jsonObjectRequest = new JsonObjectRequest(
Request.Method.POST //request method
, Constants.BASE_URL + Constants.GET_PERSON_DETAILS_URL //URL of php file
, new JSONObject(parametersMap) //parameters to send to server
, new Response.Listener<JSONObject>() { //response will come here in case of success
@Override
public void onResponse(JSONObject response) {
Log.e("response","received");
try{
txtEmail.setText(response.getString(Constants.EMAIL));
txtName.setText(response.getString(Constants.NAME));
} catch (JSONException e) {
e.printStackTrace();
}
}
},
new Response.ErrorListener() { //response will come here in case of error
@Override
public void onErrorResponse(VolleyError error) {
error.printStackTrace();
}
});
//add request to singletone
RequestSingleTone.getInstance(context).addRequest(jsonObjectRequest);
}
}
php代码:-
<?php
require_once 'dbconfig.php';
$name = json_decode(file_get_contents("php://input"),true); //sending json object from android..can not receive parameters using $_POST or $_GET
$singlePersonInfoQuery = "SELECT * FROM `test` WHERE `name`='".$name['name']."'";
$result = mysqli_query($con,$singlePersonInfoQuery);
if(mysqli_num_rows($result) > 0){
$raw = mysqli_fetch_assoc($result);
}else{
}
echo json_encode(array("name"=>$raw["name"],"email"=>$raw["email"]));
?>
这篇关于使用post方法volley将jsonobject作为请求参数传递的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
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