java - 使用HttpClient进行登录认证

Question

我正在尝试向 php 文件发送 POST 请求，当要求用户提供登录信息时，如果错误，它会打印 php 文件中的 json 消息，如果正确，它允许用户登录。但是，我的应用程序崩溃了，出现 NetworkOnThreadMainException，将错误指向三行。

 HttpResponse response=httpClient.execute(httpPost);

public class LoginActivity extends ActionBarActivity  {

login();

那么我怎样才能做到这一点呢？这是我编写的代码的一部分:

public class LoginActivity extends ActionBarActivity  {
EditText et, et2;
ImageButton ib5;
String name,pwd;

@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_login);

    et = (EditText) findViewById(R.id.editText);
    et2 = (EditText) findViewById(R.id.editText2);
    ib5 = (ImageButton) findViewById(R.id.imageButton5);

    name=et.getText().toString();
    pwd=et2.getText().toString();
    final LoginActivity loginActivity=null;

    ib5.setOnClickListener(new View.OnClickListener() {
        @Override
        public void onClick(View view) {
            //login();
            new DownloadFilesTask(loginActivity,name,pwd).doInBackground();
        }
    });

}

public void login() {
    new LoginTask(this, et.getText().toString(), et2.getText().toString());
}


private class LoginTask {
    public LoginTask(LoginActivity loginActivity, String name, String pwd) {
    }
}

void navigatetoMainActivity() {
    Intent homeIntent = new Intent(getApplicationContext(), MainActivity.class);
    homeIntent.addFlags(Intent.FLAG_ACTIVITY_CLEAR_TOP);
    startActivity(homeIntent);
}

void InvalidToast(){
    Toast.makeText(getApplicationContext(), "Please enter valid name and password", Toast.LENGTH_LONG).show();
}
void EmptyToast(){
    Toast.makeText(getApplicationContext(), "Please fill the form, don't leave any field blank", Toast.LENGTH_LONG).show();
}
}

DownloadFilesTask.java

public class DownloadFilesTask extends AsyncTask<String, String, String> {

private String name, pwd;
private LoginActivity loginActivity;

public DownloadFilesTask(LoginActivity loginActivity,String name, String pwd){
    this.loginActivity=loginActivity;
    this.name=name;
    this.pwd=pwd;
}

@Override
protected String doInBackground(String... strings) {
    HttpClient httpClient=new DefaultHttpClient();
    HttpPost httpPost=new HttpPost("login.php");
    List<NameValuePair> nameValuePairs=new ArrayList<NameValuePair>(2);
    String result=null;
    nameValuePairs.add(new BasicNameValuePair("name", name));
    nameValuePairs.add(new BasicNameValuePair("password", pwd));
    try {
        httpPost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
    } catch (UnsupportedEncodingException e) {
        e.printStackTrace();
    }
    HttpResponse response= null;
    try {
        response = httpClient.execute(httpPost);  //error is given here
    } catch (IOException e) {
        e.printStackTrace();
    }
    HttpEntity entity=response.getEntity();
    InputStream instream= null;
    try {
        instream = entity.getContent();
    } catch (IOException e) {
        e.printStackTrace();
    }
    result=convertStreamToString(instream);
    try {
        instream.close();
    } catch (IOException e) {
        e.printStackTrace();
    }



    if (Utility.isNotNull(name) && Utility.isNotNull(pwd)) {
        RequestParams params = new RequestParams();
        if (Utility.validate(name, pwd)) {
            params.put("username", name);
            params.put("password", pwd);
            onPostExecute();
        } else {
            loginActivity.InvalidToast();
        }
    } else {
        loginActivity.EmptyToast();
    }
    return result;
}


private String convertStreamToString(InputStream instream) {
    BufferedReader reader=new BufferedReader(new InputStreamReader(instream));
    StringBuilder sb=new StringBuilder();
    String line=null;
    try {
        while ((line=reader.readLine())!=null){
            sb.append(line + "\n");
        }
    } catch (IOException e) {
        e.printStackTrace();
    }finally{
        try {
            instream.close();
        } catch (IOException e) {
            e.printStackTrace();
        }
    }
    return sb.toString();
}

protected void onPostExecute(){
    loginActivity.navigatetoMainActivity();
}



private static class Utility {
    static Pattern pattern;
    static Matcher matcher;
    static Pattern pattern1;
    static Matcher matcher1;
    static String NAME_PATTERN="SuperBoise";
    static String PWD_PATTERN="qwerty";

    public static boolean validate(String name,String pwd){
        pattern=Pattern.compile(NAME_PATTERN);
        pattern1=Pattern.compile(PWD_PATTERN);
        matcher=pattern.matcher(name);
        matcher1=pattern1.matcher(pwd);
        return matcher.matches()&& matcher1.matches();
    }

    public static boolean isNotNull(String name) {
        return name!=null && name.trim().length()>0 ? true: false;
    }
}
}

Answer 1

您的应用程序有 1 个主线程在未暂停时始终运行，称为 UI 线程。

从最新版本的 Android 开始，您不允许在 UI 线程 上执行任何与网络相关的操作，因为这非常耗时，并且会阻塞负责绘制所有内容的主线程。 UI 和注册点击等(有一种方法可以绕过这个，但强烈不推荐)

执行网络相关操作(例如登录)的一种简单方法是 Android 实现的 AsyncTask 类。

该类的运行原理非常简单，它有 2 个在 UI 线程上运行的方法:onPreExecute() 和 onPostExecute() 方法。

它有一个在后台线程上运行的方法，称为doInBackground()(这是您应该执行所有网络相关操作的地方

这是 AsyncTask 类的一个非常基本的示例:

 public class DownloadFilesTask extends AsyncTask<void, void, void> {

     public DownloadFilesTask(){
          // Here you can pass data to the task
          // if you want to pass more than 1 type of data
     }

     protected void onPreExecute(){
         // this is executed on the UI Thread 
         // so you can modify elements inside the UI Thread
         // this is called after execute() is called and 
         // before doInBackground() is called
     }

     protected void doInBackground(Void... params) {
         //here you do all your network related stuff
         return null;
     }

     protected void onPostExecute(Void result) {
         // here you can work on the UI Thread
         // this is executed after the AsyncTask's execute() is finished
         // (after doInBackground() is done)
     }
 }

要使用此任务，您只需从 UI 线程 调用它，如下所示:

new DownloadFilesTask().execute();

这是developer.android.com 上的AsyncTask 文档页面:AsyncTask

您可以通过任务的构造函数传递对 LoginActivity 的引用，如果登录有效，则可以从 onPostExecute() 调用 navigatetoMainActivity() 方法> 任务内部

编辑1:您的 LoginActivity 应该是什么样子:

public class LoginActivity extends ActionBarActivity  {
EditText et, et2;
ImageButton ib5;

@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_login);

    et = (EditText) findViewById(R.id.editText);
    et2 = (EditText) findViewById(R.id.editText2);
    ib5 = (ImageButton) findViewById(R.id.imageButton5);

    ib5.setOnClickListener(new View.OnClickListener() {
        @Override
        public void onClick(View v) {
            login();
        }
    });
}

public void login(){
    new LoginTask(this, et.getText().toString(), et2.getText.toString()).execute();
}
}

编辑2:您的任务应该是这样的:

public class LoginTask extends AsyncTask<void , void, void> {

private String user, password;
private LoginActivity loginActivity;

public LoginTask(LoginActivity loginActivity, String user, String password){
     this.loginActivity = loginActivity;
     this.user = user;
     this.password = password;
}
@Override
protected String doInBackground(Void... params) {
     //do all the networking here
}

protected void onPostExecute(Void results){
     super.onPostExecute(results);
     loginActivity.navigatetoMainActivity();
}
}