Question

我想调用url，其中PHP脚本写在服务器上，在比较用户名密码后返回错误或成功但是当我点击提交按钮时它给了我这个错误

com.pace.testformwithonline E/Response error: 
com.android.volley.ParseError: org.json.JSONException: Value Error of type 
java.lang.String cannot be converted to JSONObject

package com.pace.testformwithonline;

import android.app.Activity;
import android.app.DownloadManager;
import android.support.v7.app.AppCompatActivity;
import android.os.Bundle;
import android.util.Log;
import android.view.View;
import android.widget.Button;
import android.widget.EditText;
import android.widget.TextView;
import android.widget.Toast;

import com.android.volley.Request;
import com.android.volley.RequestQueue;
import com.android.volley.Response;
import com.android.volley.VolleyError;
import com.android.volley.toolbox.JsonObjectRequest;
import com.android.volley.toolbox.Volley;

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

public class MainActivity extends Activity {
    private EditText username;
    private EditText password;
    private Button btnSubmit;
    private TextView result;
    private static final String URL="http://pace-tech.com/nomi/zaigham/android.php";
    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_main);
        username=(EditText)findViewById(R.id.editText_username);
        password= (EditText) findViewById(R.id.editText_password);
        btnSubmit=(Button)findViewById(R.id.btn_submit);
        result=(TextView)findViewById(R.id.text_result);
        btnSubmit.setOnClickListener(new View.OnClickListener() {
            @Override
            public void onClick(View v) {
                result.setText("");
            postJsonobjectRequest(URL);
            }

        });

    }
    public void postJsonobjectRequest(String url){
        JSONObject params=new JSONObject();
        RequestQueue queue = Volley.newRequestQueue(getApplicationContext());
        try {
            params.put("username", username.getText().toString());
            params.put("password", password.getText().toString());
        }
        catch (JSONException e){
         e.printStackTrace();
        }
        JsonObjectRequest jsOBJRequest=new JsonObjectRequest
                (Request.Method.POST, URL, params, new Response.Listener<JSONObject>() {

            @Override
            public void onResponse(JSONObject response) {
               // Log.e("Response is",response.toString());
                try {
                    Log.e("Response is",response.toString());
                    String Success = response.getString("Success").toString();
                    String Message = response.getString("Message").toString();
                    if(Success=="0") {
                        result.setText(Message);
                    }
                    else {
                        Toast.makeText(getApplicationContext(),Message,Toast.LENGTH_LONG);
                    }


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

            }
        }, new Response.ErrorListener() {
            @Override
            public void onErrorResponse(VolleyError error) {

                Log.e("Response error",error.toString());
                result.setText("Error Is :"+ error.toString());
            }
        });
        queue.add(jsOBJRequest);
    }
}

这是PHP SCRIPT

<?php
$username = "zaigi";
$password = "1111";

$androidUserName = json_decode($_POST['username']);
$androidPassword = json_decode($_POST['password']);

if($androidUserName == $username && $androidPassword == $password){
 $response = "Correct";
}else{
 $response = "Error";
} 
echo json_encode($response);
?>

Answer 1

这是解析错误，这意味着齐射不能解析来自服务器的响应，所以请检查响应字符串，是否采用正确的JSON格式？你可以检查你的响应字符串here，如果它是一个合适的JSON。

Answer 2

我知道这个问题。它发生在我身上一次。使用StringRequest而不是JsonObjectRequest。这是一个很大的错误。 params没有传入JsonObjectRequest。

请参阅此链接了解更多信息 Volley JsonObjectRequest Post request not working

您将获得Response as String，然后将其转换为JSONObject。

Answer 3

在PHP中SCRIPT做这样的事情..

    $data = file_get_contents('php://input');
    $json = json_decode($data);
    //username and password sent from android
    $username=$json->{'username'};
    $password=$json->{'password'};


    $androidUserName = $username;
    $androidPassword = $password;

    if($androidUserName == $myusername && $androidPassword == $mypassword){
        $response = array('result' => "Correct");
    }else{
        $response = array('result' => $androidUserName,'password' => $androidPassword);
    }
    header('Content-type: application/json');
    echo json_encode($response);

Android JSONObject请求方法POST有错误。 “无法转换为JSONObject”

3 个答案: