java.lang.String类型的值<br不能转换为JSONObject


json - Value <br of type java.lang.String cannot be converted to JSONObject

我试图登录我的应用程序,访问我的在线数据库,检查我的输入是否注册,但我得到这个错误:值

我不知道为什么php代码返回
。(我在<br表示显示)>

这是我的代码。

import android.app.Activity;
import android.app.ProgressDialog;
import android.content.Context;
import android.content.Intent;
import android.os.AsyncTask;
import android.support.v7.app.ActionBar;
import android.support.v7.app.AppCompatActivity;
import android.os.Bundle;
import android.util.Log;
import android.widget.Button;
import android.widget.EditText;
import android.view.View;
import android.widget.Toast;
import org.json.JSONObject;
import java.util.ArrayList;
import java.util.HashMap;
public class login extends Activity {
    Button btnLogIn;
    Intent intent;
    JSONObject jsonobject;   
    Button btnSignUp;
    EditText txtUsername, txtPassword;
    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_login);
        btnLogIn = (Button)findViewById(R.id.btnLogIn);
        btnSignUp = (Button)findViewById(R.id.btnSignUp);
        txtUsername = (EditText)findViewById(R.id.txtliusername);
        txtPassword = (EditText)findViewById(R.id.txtlipassword);
   //     jsonobject = JSONfunctions.getJSONfromURL("http://sql18.hostinger.ph/phpmyadmin/index.php?db=u897407316_tret&lang=en&token=6afd355a23affd65cb4d05f814cc7921&phpMyAdmin=e2ba129883c7c52bc7e30fb543d3fa1095372c90");
    }
    public void gosignup(View view){
        intent = new Intent(this, SignUp.class);
        startActivity(intent);
    }
    public void gologin(View view){
        String name = txtUsername.getText().toString();
        String pw = txtPassword.getText().toString();
        String type = "login";
        if((name.trim().equals(""))||pw.trim().equals("")){
            Toast.makeText(getApplicationContext(), "Please fill up all information", Toast.LENGTH_LONG).show();
        }
        else {
            Toast.makeText(this, "Logging in...", Toast.LENGTH_SHORT).show();
            new LogInAction(this).execute(name, pw);
           // intent = new Intent(this, UserHomeDrawer.class);
            //startActivity(intent);
        }
    }
}

LogInAction.java

import android.content.Context;
import android.content.Intent;
import android.os.AsyncTask;
import android.util.Log;
import android.widget.Toast;
import org.json.JSONException;
import org.json.JSONObject;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.UnsupportedEncodingException;
import java.net.HttpURLConnection;
import java.net.MalformedURLException;
import java.net.URL;
import java.net.URLEncoder;
/**
 * Created by Camille on 11/13/2016.
 */
public class LogInAction extends AsyncTask<String, Void, String> {
    private Context context;
    public LogInAction(Context context) {
        this.context = context;
    }
    protected void onPreExecute() {
    }
    @Override
    protected String doInBackground(String... arg0) {
        String username = arg0[0];
        String password = arg0[1];
        String link;
        String data;
        BufferedReader bufferedReader;
        String result;
        try {
            data = "?username=" + URLEncoder.encode(username, "UTF-8");
            data += "&password=" + URLEncoder.encode(password, "UTF-8");
            link = "http://threatmam.esy.es/loginActionAndroid.php" + data;
            URL url = new URL(link);
            HttpURLConnection con = (HttpURLConnection) url.openConnection();
            bufferedReader = new BufferedReader(new InputStreamReader(con.getInputStream()));
            result = bufferedReader.readLine();
           return result;
        } catch (Exception e) {
            return new String("Exception: " + e.getMessage());
        }
    }
    @Override
    protected void onPostExecute(String result) {
        String jsonStr = result.toString();
        if (jsonStr != null) {
            try {
                JSONObject jsonObj = new JSONObject(jsonStr);
                //  JSONObject jsonObj = new JSONObject(jsonStr.substring(jsonStr.indexOf("{"), jsonStr.lastIndexOf("}") + 1));
                String query_result = jsonObj.getString("query_result");
                if (query_result.equals("SUCCESS")) {
                    Toast.makeText(context, "Log in successful!", Toast.LENGTH_SHORT).show();
                    Intent intent = new Intent(context,UserHomeDrawer.class)
                            .setFlags(Intent.FLAG_ACTIVITY_NEW_TASK);
                    context.startActivity(intent);
                } else if (query_result.equals("FAILURE")) {
                    Toast.makeText(context, "Log in failed.", Toast.LENGTH_SHORT).show();
                } else {
                    Toast.makeText(context, "Couldn't connect to remote database.", Toast.LENGTH_SHORT).show();
                }
            } catch (JSONException e) {
                e.printStackTrace();
                Log.d("JSON", "Error:",e);
                Toast.makeText(context, ""+e+"RES: "+jsonStr, Toast.LENGTH_SHORT).show();
            }
        } else {
            Toast.makeText(context, "Couldn't get any JSON data.", Toast.LENGTH_SHORT).show();
        }
    }
}
这是我的PHP代码
<?php
$un = $_GET['username'];
$pw = md5(sha1($_GET['password']));
$res = 0;
include "connect.php";

$query = "SELECT * FROM tbl_mobileuser";
result = $mysqli->query($query);
if ($result->num_rows > 0) {
    while($row = $result->fetch_assoc()) {
        if(($un == $row['Username']) && if($pw == $row['Password'])){
            echo '{"query_result":"SUCCESS"}';
            $res = 1;
        }
    }
}
else{
    echo '{"query_result":"FAILURE"}';
}

if($res == 0){
    echo '{"query_result":"FAILURE"}';
}                       
?>

我确实查找了可能的解决方案,但我仍然无法解决这个问题。我希望有人能帮助我。

你好像在第11行漏掉了一个$

改变,

result = $mysqli->query($query);

$result = $mysqli->query($query);

注意:您没有正确读取php脚本的响应。readLine()只读取一行。如果你的php脚本输出多个(就像现在一样),你的android代码将无法读取它,你会丢失一些数据。

我点击你的URL如下:

http://threatmam.esy.es/loginActionAndroid.php?username=foo&密码=酒吧

服务器的响应是:

<br /> <b>Parse error</b>: syntax error, unexpected '=' in <b>/home/u897407316/public_html/loginActionAndroid.php</b> on line <b>10</b><br />

看起来你有你的答案就在那里-请求在服务器端失败,并返回一个错误响应,其中包含HTML,你试图将其转换为JSON。很自然,这是行不通的。

确保你向服务器发送了正确的请求,并且服务器代码正常工作。

此外,您可能会考虑使用现有的优秀HTTP请求库之一(Volley, OkHttp, Retrofit等),而不是使用HttpURLConnection推出自己的解决方案。