1

我从 android 开始（3 天前），我无法得到我想要的解决方案。所以，我在这里读了很多关于 asyncTask 的帖子，现在我确定我很困惑。

首先，这是我的第一个问题，所以我希望我至少能做到这一点。

我想要的是有一个类来连接到某个服务器以及它的结果。之后分析 json 或 xml。

所以这就是我所做的。这是我的活动课（从主课调用）

    public class LogIn extends Activity implements OnClickListener {

    Button btn =null;

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

    btn = (Button) findViewById(R.id.btn_login);
    btn.setOnClickListener( this);
}   

public void onClick(View view) {
    HttpResponse response;
    Intent data = new Intent();
    //---get the EditText view---
    EditText txt_user = (EditText) findViewById(R.id.et_un);
    EditText txt_pwd =  (EditText) findViewById(R.id.et_pw);

    // aca tengo q llamar al conectar y chequear en BD
    ArrayList<NameValuePair> postParameters = new ArrayList<NameValuePair>();
    postParameters.add(new BasicNameValuePair("usr", txt_user.getText().toString()));
    postParameters.add(new BasicNameValuePair("pass", txt_pwd.getText().toString()));
    String URL="whatever";
    try {
        response= new ConectServer(URL, postParameters).execute().get();

        LeerAutentificacionXml l= new LeerAutentificacionXml(response);
        String s=l.Transformar();
        data.setData(Uri.parse(s.toString()));
        setResult(RESULT_OK, data);
    } catch (InterruptedException e) {
        e.printStackTrace();
    } catch (ExecutionException e) {
        e.printStackTrace();
    } catch (IllegalStateException e) {
        e.printStackTrace();
    } catch (IOException e) {
        e.printStackTrace();
    }

    //---set the data to pass back---
//  data.setData(Uri.parse(txt_user.getText().toString()+ " " + Uri.parse(txt_pwd.getText().toString())));
//  setResult(RESULT_OK, data);
    //---closes the activity---
    finish();
} }

这是我连接到网络服务的课程。

    public class ConectServer extends AsyncTask <Void, Void, HttpResponse> {

private String URL=null;
private ArrayList<NameValuePair> postParameters=null;
/** Single instance of our HttpClient */
private HttpClient mHttpClient;
/** The time it takes for our client to timeout */
public static final int HTTP_TIMEOUT = 30 * 1000;


  public ConectServer(String url, ArrayList<NameValuePair> p) {
      this.URL=url;
      this.postParameters=p;
  }

 private HttpClient getHttpClient() {
     if (mHttpClient == null) {
         mHttpClient = new DefaultHttpClient();
         final HttpParams params = mHttpClient.getParams();
         HttpConnectionParams.setConnectionTimeout(params, HTTP_TIMEOUT);
         HttpConnectionParams.setSoTimeout(params, HTTP_TIMEOUT);
         ConnManagerParams.setTimeout(params, HTTP_TIMEOUT);
     }
     return mHttpClient;
 }

 /**
 * Performs an HTTP Post request to the specified url with the
 * specified parameters.
 *
 * @param url The web address to post the request to
 * @param postParameters The parameters to send via the request
 * @return The result of the request
 * @throws Exception
 */
@Override
protected HttpResponse doInBackground(Void... params) {
    // TODO Auto-generated method stub
     HttpResponse response = null;
     try {
         HttpClient client = getHttpClient();
         HttpPost request = new HttpPost(this.URL);
         UrlEncodedFormEntity formEntity;
         formEntity = new UrlEncodedFormEntity(this.postParameters);
         request.setEntity(formEntity);
         response = client.execute(request);
     } catch (UnsupportedEncodingException e) {
         e.printStackTrace();
     } catch (ClientProtocolException e) {
        e.printStackTrace();
     } catch (IOException e) {
        e.printStackTrace();
    }

    return response;
}


public void onPreExecute() {
     super.onPreExecute();

}

protected void onPostExecute(HttpResponse response) {
     super.onPostExecute(response);
}}

我读到了一些关于听众的设计模式，但首先我想更好地理解为什么这不起作用。我从服务器收到一个错误，我想知道这是否正确或发生了哪个大的 newby 失败。

提前谢谢。

4

2 回答 2

2

调用asyncTask.get()等待AsyncTask完成，然后返回结果。它基本上违背了目的AsyncTask- 关键AsyncTask是长时间运行的任务在后台线程中执行，并且在此期间 UI 线程没有被阻塞。当您调用.get()它时，它会阻止 UI 线程等待后台线程完成。

所以，不要使用get()和移动当结果可用时应该发生的所有动作onPostExecute(..)。像这样的东西：

protected void onPostExecute(HttpResponse response) {
    super.onPostExecute(response);
    LeerAutentificacionXml l= new LeerAutentificacionXml(response);
    String s=l.Transformar();
    data.setData(Uri.parse(s.toString()));
}

当然，您必须将一些引用（data？）传递给AsyncTask（通过构造函数或其他方式）。

于 2011-09-26T21:48:18.670 回答
1

尝试将所有网络代码封装在异步任务中，这样您就可以这样调用它：

EditText txt_user = (EditText) findViewById(R.id.et_un);
EditText txt_pwd =  (EditText) findViewById(R.id.et_pw);
new LoginTask(txt_user.getText(), txt_pwd.getText(), this).execute();

根据LoginTask您问题中的代码，类似于以下内容：

public class LoginTask extends AsyncTask<Void, Void, String> {

    // The URL probably won't change, so keep it in a static field
    private final static String URL = "http://....";

    private final String username;
    private final String password;
    private final Activity activity;

    /*
     * Pass all data required to log in and handle the result here.
     */
    public LoginTask(final String username, final String password, final Activity activity) {
        this.username = username;
        this.password = password;
        this.activity = activity;
    }

    /*
     * Do all the network IO and time-consuming parsing in here,
     * this method will be invoked in its own thread, not blocking the UI.
     */
    @Override
    protected String doInBackground(Void... params) {
        HttpClient client = getHttpClient();
        try {
            HttpEntity formEntity = new UrlEncodedFormEntity(Arrays.asList(
                    new BasicNameValuePair("usr", username),
                    new BasicNameValuePair("pass", password)));

            HttpPost request = new HttpPost(URL);
            request.setEntity(formEntity);

            HttpResponse response = client.execute(request);

            LeerAutentificacionXml l= new LeerAutentificacionXml(response);
            return l.Transformar();
        } catch (IOException ex) {
            // properly log your exception, don't just printStackTrace()
            // as you can't pinpoint the location and it might be cut off
            Log.e("LOGIN", "failed to log in", ex);
        } finally {
            client.getConnectionManager().shutdown();
        }
    }

    private HttpClient getHttpClient() {
        return null; // insert your previous code here
    }

    /*
     * This again runs on the UI thread, so only do here
     * what really needs to run on the UI thread.
     */
    @Override
    protected void onPostExecute(String response) {
        Intent data = new Intent();
        data.setData(Uri.parse(response));
        activity.setResult(Activity.RESULT_OK, data);
    }
}

即使您不认为登录是一项“长期运行的任务”，任何涉及网络 IO 的定义都是这样的任务，因为它可能由于您无法控制的原因而挂起、超时等。

有多个类做网络 IO 并没有错，只要像上面的LoginTask例子那样仔细封装细节。试着写两三个AsyncTask这种类型的类，然后看看你能不能把共同的部分提取成一个共同的“网络东西” AsyncTask。但是，不要一次全部完成。

于 2011-09-26T22:04:08.250 回答