Android匿名asyncTask方法中的返回值

Question

现在在我的应用程序中，我尝试从 url 进行 http 解析，但在此之前我没有携带线程...

我有这样的类和方法：

public class TwitterOAuthHelper {
public String httpQueryToApi(String url) {
        HttpGet get = new HttpGet(url);
        HttpParams params = new BasicHttpParams();
        HttpProtocolParams.setUseExpectContinue(params, false);
        get.setParams(params);
        String response = null;
        try {
            SharedPreferences settings = context.getSharedPreferences("my_app", 0);
            String userKey = settings.getString("user_key", "");
            String userSecret = settings.getString("user_secret", "");
            consumer.setTokenWithSecret(userKey, userSecret);
            consumer.sign(get);
            DefaultHttpClient client = new DefaultHttpClient();
            response = client.execute(get, new BasicResponseHandler());
        } catch (Exception e) {
            displayToast("Failed to get data.");
        }
        return response;
    }

}

现在我尝试将此逻辑移动到 asyncTask 中：

String result;
public String httpQueryToApi(String url) {
    new AsyncTask<String,Void,String>(){

        @Override
        protected String doInBackground(String... params) {
            HttpGet get = new HttpGet(String.valueOf(params));
            HttpParams param = new BasicHttpParams();
            HttpProtocolParams.setUseExpectContinue(param, false);
            get.setParams(param);
            String response = null;
            try {
                SharedPreferences settings = context.getSharedPreferences("my_app", 0);
                String userKey = settings.getString("user_key", "");
                String userSecret = settings.getString("user_secret", "");
                consumer.setTokenWithSecret(userKey, userSecret);
                consumer.sign(get);
                DefaultHttpClient client = new DefaultHttpClient();
                response = client.execute(get, new BasicResponseHandler());
            } catch (Exception e) {
                displayToast("Failed to get data.");
            }
            result = response;
            return response;
        }
    }.execute(url);
    return result;
    }

但是如何将我的响应结果值返回给方法？

这样做的最佳做法是什么？

Answer 1

将下面的方法添加到异步任务主体（doInBackground 方法下面）：

@Override
protected void onPostExecute(String result) {
    // result is your returned value from doInBackground
    // now we are in main ui thread
}

如果你想回调另一个方法，它应该是接口

public interface ResultInterface {
     public void resultFromHttp(String result);
}

然后你的方法

public String httpQueryToApi(String url, final ResultInterface ri){
      //as bove
    @Override
    protected void onPostExecute(String result) {
        if(ri!=null)
            ri.resultFromHttp(result);
    }

}

在您的Activity/Fragment/调用httpQueryToApi 中实现ResultInterface，将this 作为第二个参数传递（ri 接口）

Answer 2

你不能，因为任务将在另一个线程中执行。

但是，您可以使用回调来获取结果。 看看：https://stackoverflow.com/a/19520293/4299154.

Answer 3

最初，您的函数返回一个字符串，然后您对其进行进一步处理。美好的。 但是你不能像那样使用线程。你不能从函数返回结果，因为它还没有被设置（你可以，但它会返回 null）。这样做的正确方法是

public void httpQueryToApi(String url) {

    new AsyncTask<String,Void,String>(){

        @Override
        protected String doInBackground(String... params) {
            HttpGet get = new HttpGet(String.valueOf(params));
            HttpParams param = new BasicHttpParams();
            HttpProtocolParams.setUseExpectContinue(param, false);
            get.setParams(param);
            String response = null;
            try {
                SharedPreferences settings = context.getSharedPreferences("my_app", 0);
                String userKey = settings.getString("user_key", "");
                String userSecret = settings.getString("user_secret", "");
                consumer.setTokenWithSecret(userKey, userSecret);
                consumer.sign(get);
                DefaultHttpClient client = new DefaultHttpClient();
                response = client.execute(get, new BasicResponseHandler());
            } catch (Exception e) {
                displayToast("Failed to get data.");
            }
            return response;
        }

        @Override
        protected void onPostExecute(String s) {

            //here s is the response string, do what ever you want
            super.onPostExecute(s);
        }
    }.execute(url);              
}

您将不得不将您的进一步处理逻辑转移到 onPostExecute，别无他法:) 如果你想深入了解Future<>

Answer 4

1) 为您创建单独的异步类（非匿名）。

2) 创建接口类。

public interface AsyncResponse {
    void onProcessFinish(String output);
}

3) 在你的 Async 类中，你需要声明它（接口：AsyncResponse）：

public class MyAsyncTask extends AsyncTask<String,Void,String>(){
   public AsyncResponse listener = null;    

   public MyAsyncTask(AsyncResponse l) {
        this.listener = l;
    }
   {...}
   @Override
        protected String doInBackground(String... params) {
            HttpGet get = new HttpGet(String.valueOf(params));
            HttpParams param = new BasicHttpParams();
            HttpProtocolParams.setUseExpectContinue(param, false);
            get.setParams(param);
            String response = null;
            try {
                SharedPreferences settings = context.getSharedPreferences("my_app", 0);
                String userKey = settings.getString("user_key", "");
                String userSecret = settings.getString("user_secret", "");
                consumer.setTokenWithSecret(userKey, userSecret);
                consumer.sign(get);
                DefaultHttpClient client = new DefaultHttpClient();
                response = client.execute(get, new BasicResponseHandler());
            } catch (Exception e) {
                displayToast("Failed to get data.");
            }
            result = response;
            return response;
        }


   @Override
   protected void onPostExecute(String result) {
      listener.onProcessFinish(result);
   }
}

4) 在您的类中（您在其中调用 AssyncClass，例如在您的 Activity 中）您需要实现您之前创建的 AsyncResponse 接口。

public class MainActivity implements AsyncResponse{



  {...}
   void onProcessFinish(String output){
     //this you will received result fired from async class of onPostExecute(result) method.
   }
}

6) 现在您可以在 MainActivity 中调用：

new MyAsyncTask(this).execute("your_url");