我想登錄我的應用程序,訪問我的在線數據庫,並檢查我的輸入是否已註冊,但我得到此錯誤:Value <類型爲java.lang.String的br不能轉換爲JSONObject。json - 值類型java.lang.String不能轉換爲JSONObject
我不知道爲什麼我的php代碼返回< br />。 (我增加了<和br之間的空格以顯示)
這是我的代碼。
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();
}
}
}
這是我的PHP代碼
<?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"}';
}
?>
我絕對擡頭可能的解決辦法,但我仍然無法解決這個問題。我希望有一個人可以幫助我。
'Log.d(「JSON」,jsonStr)'...調試請求,您將清楚地看到問題 –
運行php腳本。這是輸出。 _
解析錯誤:語法錯誤,意想不到的 '=' 在loginActionAndroid.php上線
_ –