使用AsyncTask登录用户

Question

我需要获取user并将其保存到变量。

public class MainActivity extends ActionBarActivity 
{
    User user = new LoginTask2().execute("");
}

class LoginTask2 extends AsyncTask<String, Void, User> {
    private Exception exception;
    public String hash = "";

    protected String doInBackground(String... t) {
        RestClient restClient = new HttpRestClient();
        restClient.setUserAgent("bot/1.0 by name");

        // Connect the user
        User user = new User(restClient, "User", "somepass");
        try {
            user.connect();
            //hash = user.getModhash();
            return user;
        } catch (Exception e) {
            e.printStackTrace();
            this.exception = e;
            return null;
        }
    }

    protected void onPostExecute(String string) {

    }
}

看起来不错，但是我不知道该如何获取user 。 有了这段代码，我得到错误：

Error:(49, 47) error: incompatible types
required: String
found:    AsyncTask<String,Void,String>

有人可以给我建议如何更改代码吗？

Answer 1

您的异步任务已声明：

class LoginTask2 extends AsyncTask<String, Void, User> {
    protected String doInBackground(String... t) {

为了避免不兼容类型错误，您需要使doInbackground返回一个User对象。

    protected User doInBackground(String... t) {

请参阅： http : //developer.android.com/reference/android/os/AsyncTask.html

您上面的代码：

User user = new LoginTask2().execute("");

也会失败，因为您必须执行异步任务，然后再使用返回值。 您可以通过将返回的User对象设置为MainActivity一个字段来访问它，然后在AsyncTask完成后再使用该对象。

Answer 2

将asyntask定义为内部类，onPostExecute分配mUser类变量。

public class MainActivity extends ActionBarActivity 
    {
        User mUser;

        new LoginTask2().execute("");

    class LoginTask2 extends AsyncTask<String, Void, User> {
        private Exception exception;
        public String hash = "";

        protected User doInBackground(String... t) {
            RestClient restClient = new HttpRestClient();
            restClient.setUserAgent("bot/1.0 by name");

            // Connect the user
            User user = new User(restClient, "User", "somepass");
            try {
                user.connect();
                //hash = user.getModhash();
                return user;
            } catch (Exception e) {
                e.printStackTrace();
                this.exception = e;
                return null;
            }
        }

        protected void onPostExecute(User user) {
                mUser = user;
        }
    }
    }

Answer 3

调用execute方法时，AsyncTask将不会返回用户变量。 因此，以下代码将不起作用。

User user = new LoginTask2().execute("");

让我们进行一些更改。

private class LoginTask2 extends AsyncTask<String, Void, User> {
    private Exception exception;
    public String hash = "";

    @Override
    protected String doInBackground(String... t) {
        RestClient restClient = new HttpRestClient();
        restClient.setUserAgent("bot/1.0 by name");

        // Connect the user
        User user = new User(restClient, "User", "somepass");
        try {
            user.connect();
            //hash = user.getModhash();
            return user;
        } catch (Exception e) {
            e.printStackTrace();
            this.exception = e;
            return null;
        }
    }

    protected void onPostExecute(User user) {

    }
} 

由于AsyncTask仅属于此类，因此应将其设为私有。 方法doInBackground的返回值也是onPostExecute的参数。

为了保存用户数据，您可以采取以下几种方法：

可以使用onPostExecute方法保存您的数据

protected void onPostExecute(User user) {
     //do save stuff
}

您还可以从类中调用一个方法，例如：

public class MainActivity extends ActionBarActivity {
    User user = new LoginTask2().execute("");

    private void success(User user){
       //do save stuff
    }

    private void failure(){

    } 

    private class LoginTask2 extends AsyncTask<String, Void, User> {
        private Exception exception;
        public String hash = "";

        @Override
        protected String doInBackground(String... t) {
            RestClient restClient = new HttpRestClient();
            restClient.setUserAgent("bot/1.0 by name");

            // Connect the user
            User user = new User(restClient, "User", "somepass");
            try {
                user.connect();
               //hash = user.getModhash();
               return user;
            } catch (Exception e) {
               e.printStackTrace();
               this.exception = e;
               return null;
           }
       }

       protected void onPostExecute(User user) {
             if(user != null)
                success(user)
             else
                failure()
       }
    } 
}

您也可以遇到失败:)