Android與MySQL登錄頁​​面

Question

嘿，我是新來的Android，我有麻煩登錄頁面。我試圖連接到包含用戶名和密碼的MySQL數據庫。我試圖讓我的代碼工作，但它只是不會讓我登錄。

這裏是我的代碼

package login.CBA; 

import android.app.Activity; 
import android.content.Intent; 
import android.os.Bundle; 
import android.os.StrictMode; 
import android.view.View; 
import android.view.View.OnClickListener; 
import android.widget.Button; 
import android.widget.EditText; 
import android.widget.TextView; 

import java.util.ArrayList; 

import org.apache.http.NameValuePair; 
import org.apache.http.message.BasicNameValuePair; 
import org.apache.http.params.HttpProtocolParams; 

public class login extends Activity { 

private EditText etUsername; 

private EditText etPassword; 

private Button btnLogin; 

private Button btnCancel; 

private TextView lblResult; 


@Override 


public void onCreate(Bundle savedInstanceState) { 

    super.onCreate(savedInstanceState); 



    setContentView(R.layout.main); 



    etUsername = (EditText)findViewById(R.id.Username); 

    etPassword = (EditText)findViewById(R.id.password); 

    btnLogin = (Button)findViewById(R.id.login); 

    btnCancel = (Button)findViewById(R.id.Cancel); 

    lblResult = (TextView)findViewById(R.id.result); 



    btnLogin.setOnClickListener(new OnClickListener() { 

     @Override 

     public void onClick(View v) { 



      StrictMode.ThreadPolicy policy = new                    
      StrictMode.ThreadPolicy.Builder().permitAll().build(); 
      StrictMode.setThreadPolicy(policy); 



      ArrayList<NameValuePair> postParameters = new ArrayList<NameValuePair>(); 

      postParameters.add(new BasicNameValuePair("Username", 
      etUsername.getText().toString())); 

      postParameters.add(new BasicNameValuePair("password", 
      etPassword.getText().toString())); 

      /*   String valid = "1";*/  

      String response = null; 


      try { 

       response = 
       CustomHttpClient.executeHttpPost("localhost/check.php", 
       postParameters); 

       String res = response.toString(); 

       //res = res.trim(); 

       res = res.replaceAll("\\s+",""); 

       // error.setText(res); 

       if (res.equals("1")) { 

        Intent i = new Intent(getApplicationContext(), mainmenu.class); 
       startActivity(i); 

       } 

       else lblResult.setText("Incorrect Username or Password Entered"); 

      } 

      catch (Exception e) { 

       etUsername.setText(e.toString()); 


      } 



     } 


    }); 
    } 
} 

和PHP代碼

<?php 

$un=$_POST['username']; 
$pw=$_POST['password']; 

$conn = mysql_connect('localhost', 'root', 'user'); 
mysql_select_db('cba'); 

$query = 'SELECT * FROM login WHERE username = ‘$un’ AND password = ‘$pw’'; 
$result = mysql_query($query) or die ('Unable to verify user because : ' .  
mysql_error()); 

if (mysql_num_rows($result) > 0) { 

echo 1; 
} 
else { 

// print status message 
echo 0; 
} 
?> 

Answer 1

我認爲這條線上可能存在問題：

$query = 'SELECT * FROM login WHERE username = ‘$un’ AND password = ‘$pw’'; 

看起來你正在嘗試使用不同種類的單引號。試試這個：

$query = "SELECT * FROM login WHERE username = '$un' AND password = '$pw'"; 

此外，你應該逃避用戶名和密碼的特殊字符，以防止SQL注入攻擊。

$un = mysql_real_escape_string($_POST['username']); 
$pw = mysql_real_escape_string($_POST['password']); 

退房過去有關信息SO問題 「org.apache.http.NoHttpResponseException：目標服務器無法響應」HttpClient on Android : NoHttpResponseException through UMTS/3G

Answer 2

您需要更改這個

postParameters.add(new BasicNameValuePair("Username", 

進入此

postParameters.add(new BasicNameValuePair("username", 

字符串必須相同

Answer 3

我想也許你忘了將活動添加到AndroidManifest.xml文件中。

只需添加：

activity android:name=".mainmenu" 

到AndroidManifest.xml中