从Android发送数据到php webservice验证登录

Question

我有一个login.php脚本，它将验证在android中输入的用户名和密码。代码是

<?php 

    include('dbconnect.php'); 

    $data=file_get_contents('php://input'); 
    $json = json_decode($data); 
    $tablename = "users"; 

    //username and password sent from android 
    $username=$json->{'username'}; 
    $password=$json->{'password'}; 

    //protecting mysql injection 
    $username = stripslashes($username); 
    $password = stripslashes($password); 
    $username = mysql_real_escape_string($username); 
    $password = mysql_real_escape_string($password); 
    $password = md5($password); 

    $sql = "SELECT id FROM $tablename WHERE u_username='$username' and password='$password'"; 

    //Querying the database 
    $result=mysql_query($sql); 

    //If found, number of rows must be 1 
    if((mysql_num_rows($result))==1){ 

    //creating session 
    session_register("$username"); 
    session_register("$password"); 

    print "success"; 
    }else{ 
    print "Incorrect details"; 
    } 
?> 

我也有一个android类，用户将从中输入用户名和密码。代码如下。

public class LoginActivity extends Activity { 


    public static final String loginURI="http://.../login.php"; 

    @Override 
    public void onCreate(Bundle savedInstanceState){ 
    super.onCreate(savedInstanceState); 
    setContentView(R.layout.login); 



    buttonSubmit.setOnClickListener(new View.OnClickListener() { 
     @Override 
     public void onClick(View view) { 


      String userID = ""; 
      userID=login(editTextUsername.getText().toString(), editTextPassword.getText().toString()); 

      if (editTextPassword.getText().toString() != null & editTextUsername.getText().toString() != null){ 
       //Used to move to the Cases Activity 
       Intent casesActivity = new Intent(getApplicationContext(), CasesActivity.class); 
       startActivity(casesActivity); 
       casesActivity.putExtra("username", userID); 

      } 
      else{ 
       //Display Toaster for error 
       Toast.makeText(getApplicationContext(),"this is an error message", Toast.LENGTH_LONG).show(); 
      } 
     } 
    }); 


private String login(String username, String password){ 

    JSONObject jsonObject = new JSONObject(); 

    String success = ""; 

    HttpClient httpClient = new DefaultHttpClient(); 
    HttpPost httpPost = new HttpPost(loginURI); 
    HttpParams httpParams = new BasicHttpParams(); 
    HttpConnectionParams.setConnectionTimeout(httpParams,10000); 
    HttpConnectionParams.setSoTimeout(httpParams,10000); 

    try { 

     jsonObject.put("username", username); 
     Log.i("username", jsonObject.toString()); 
     jsonObject.put("password", password); 
     Log.i("password", jsonObject.toString()); 

     StringEntity stringEntity = new StringEntity(jsonObject.toString()); 
     stringEntity.setContentEncoding(new BasicHeader(HTTP.CONTENT_TYPE, "application/json")); 
     httpPost.setEntity(stringEntity); 
     HttpResponse httpResponse = httpClient.execute(httpPost); 

     HttpEntity entity = httpResponse.getEntity(); 

     if (entity != null) { 
      success = EntityUtils.toString(httpResponse.getEntity()); 
      Log.i("success", success); 
     } 

    }catch (IOException e){ 
     Log.e("Login_Issue", e.toString()); 
    }catch (JSONException e) { 
     e.printStackTrace(); 
    } 


    return success; 
    } 

} 

我收到以下错误：错误/ AndroidRuntime（29611）：致命异常：主要android.os.NetworkOnMainThreadException。 我知道我需要另一个线程，我正在考虑使用AsyncTask，但我不知道该把它放在这个类中。

你可以给我一些建议，使用JSON发送和接收来自android的数据。

感谢你的帮助，

Answer 1

您可以通过调用内部doInBackground登录方法更改使用的AsyncTask代码和onPostExecute开始下一个活动时，登录成功，因为：

private class LoginOperation extends AsyncTask<String, Void, String> { 

String str_username=; 
String str_password=; 

    public LoginOperation(String str_username,String str_password){ 
     this.str_password= str_password; 
     this.str_username= str_username; 
    } 
    @Override 
     protected void onPreExecute() { 

     // show progress bar here 
     } 
     @Override 
     protected String doInBackground(String... params) { 
      // call login method here 
     String userID=login(str_username,str_password); 
     return userID; 
     }  

     @Override 
     protected void onPostExecute(String result) {  
     // start next Activity here 
      if(result !=null){ 
       Intent casesActivity = new Intent(getApplicationContext(), 
               CasesActivity.class); 
       casesActivity.putExtra("username", result); 
       Your_Activiy.this.startActivity(casesActivity); 

       } 
     } 

，并开始LoginOperation的AsyncTask上按钮点击为：

buttonSubmit.setOnClickListener(new View.OnClickListener() { 
     @Override 
     public void onClick(View view) { 

      if (editTextPassword.getText().toString() != null 
         & editTextUsername.getText().toString() != null){ 
       // start AsyncTask here 
       new LoginOperation(editTextUsername.getText().toString(), 
         editTextPassword.getText().toString()).execute(""); 
      } 
      else{ 
       // your code here 
      } 
     } 
    }); 

} 

Answer 2

简单的答案是创建一个线程，并只在该t内调用登录hread或Async任务（您可以将其定义为一个新类，然后调用execute）。就像这样：

旧代码：

userID=login(editTextUsername.getText().toString(), editTextPassword.getText().toString()); 

新代码：

Runnable runnable = new Runnable() { 
     void run() { 
     login(editTextUsername.getText().toString(), editTextPassword.getText().toString()); 
     } 
(new Thread(runnable)).start();