Question

我正在使用具有以下签名的AsyncTask类：

public class ApiAccess extends AsyncTask<List<NameValuePair>, Integer, String> {
    ...
private String POST(List<NameValuePair>[] nameValuePairs){
    ...
    return response;
}
}

protected String doInBackground(List<NameValuePair>... nameValuePairs) {
    return POST(params);
}

我试图通过以下方式从其他类中调用它：

ApiAccess apiObj = new ApiAccess (0, "/User");
// String signupResponse = apiObj.execute(nameValuePairs);
String serverResponse = apiObj.execute(nameValuePairs); //ERROR

但是我在这里得到了这个错误：

Type mismatch: cannot convert from AsyncTask<List<NameValuePair>,Integer,String> to String

为什么我在类扩展行中指定String作为第三个参数？

Answer 1

您可以通过在返回的AsyncTask上调用AsyhncTask的get（）方法来获得结果，但是当它等待获得结果时，它会将它从异步任务转换为同步任务。

String serverResponse = apiObj.execute(nameValuePairs).get();

由于您将AsyncTask放在一个单独的类中，您可以创建一个接口类并在AsyncTask中声明它，并在您希望从中访问结果的类中将您的新接口类实现为委托。这里有一个很好的指南：How to get the result of OnPostExecute() to main activity because AsyncTask is a separate class?。

我将尝试将上述链接应用于您的上下文。

（IApiAccessResponse）

public interface IApiAccessResponse {
    void postResult(String asyncresult);
}

（ApiAccess）

public class ApiAccess extends AsyncTask<List<NameValuePair>, Integer, String> {
...
    public IApiAccessResponse delegate=null;
    protected String doInBackground(List<NameValuePair>... nameValuePairs) {
        //do all your background manipulation and return a String response
        return response
    }

    @Override
    protected void onPostExecute(String result) {
        if(delegate!=null)
        {
            delegate.postResult(result);
        }
        else
        {
            Log.e("ApiAccess", "You have not assigned IApiAccessResponse delegate");
        }
    } 
}

（您的主要类，实现IApiAccessResponse）

ApiAccess apiObj = new ApiAccess (0, "/User");
//Assign the AsyncTask's delegate to your class's context (this links your asynctask and this class together)
apiObj.delegate = this;
apiObj.execute(nameValuePairs); //ERROR

//this method has to be implement so that the results can be called to this class
void postResult(String asyncresult){
     //This method will get call as soon as your AsyncTask is complete. asyncresult will be your result.
}

Answer 2

我建议实现一个Handler Callback。您可以将片段的（或活动的）处理程序传递给AsyncTask，AsyncTask在完成时将调用它。 AsyncTask也可以传回一个任意对象。

这是一个AsyncTask的例子，我在它自己的文件中（没有子类）：

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

    private static final String TAG = "MyTask";
    private Handler mCallersHandler;
    private Candy    mObject1;
    private Popsicle mObject2;

    // Return codes
    public static final int MSG_FINISHED = 1001;

    public SaveVideoTask(Handler handler, Candy candyCane, Popsicle grapePop ) {
        this.mCallersHandler = handler;
        this.mObject1        = candyCane;
        this.mObject2        = grapePop;
    }

    @Override
    protected String doInBackground(Void... params) {

        // Do all of the processing that you want to do...
        // You already have the private fields because of the constructor
        // so you can use mObject1 and mObject2
        Dessert objectToReturn = mObject1 + mObject2;

        // Tell the handler (usually from the calling thread) that we are finished, 
        // returning an object with the message
        mCallersHandler.sendMessage( Message.obtain( mCallersHandler, MSG_FINISHED, objectToReturn ) );

        return (null);
    }
}

此示例假设您的AsyncTask需要一块Candy和一个冰棒。然后它会将一个甜点返回到你的片段。

您可以使用以下代码在片段的一行中构建和运行AsyncTask：

( new MyTask( mFragmentHandler, candyCane, grapePop ) ).execute();

但是，当然，您首先需要设置片段的处理程序（myFragmentHandler）。为此，您的片段（或活动）应该看起来像（注意＆＃34;实现Handler.Callback＆＃34;）：

public class MyFragment extends Fragment implements Handler.Callback {

    private Handler mFragmentHandler;
    private Candy candyCane;
    private Popsicle grapePop;

    @Override
    public void onCreate(Bundle savedInstanceState) {

        // Standard creation code
        super.onCreate(savedInstanceState);
        setRetainInstance(true);

        // Create a handler for this fragment 
        mFragmentHandler = new Handler(this);

        // Other stuff...
    }

    @Override
    public View onCreateView(LayoutInflater inflater, ViewGroup parent,
            Bundle savedInstanceState) {

        // Inflate the layout
        View v = inflater.inflate(R.layout.my_fragment_layout, parent, false );

        // The candyCane and grapePop don't need to be set up here, but 
        // they MUST be set up before the button is pressed. 
        // Here would be a good place to at least initialize them...

        // Perhaps you have a button in "my_fragment_layout" that triggers the AsyncTask...
        Button mButton  = (Button) v.findViewById(R.id.mButton);
        mButton.setOnClickListener( new OnClickListener() {
            @Override
            public void onClick(View v) {
                ( new MyTask( mFragmentHandler, candyCane, grapePop ) ).execute();
            }
        });
        return v;
    }

    @SuppressWarnings("unchecked")
    @Override
    public boolean handleMessage(Message msg) {

        switch (msg.what) {

        case MyTask.MSG_FINISHED:

            // Let's see what we are having for dessert 
            Dessert myDessert = (Dessert) msg.obj;
            break;

        }
        return false;
    }

}

如果您使用这些代码，按下按钮将触发AsyncTask。在AsyncTask处理时，调用片段将继续执行。然后，当AsyncTask完成时，它将向片段发送一条消息，告知它已完成，并传递带有该消息的对象。此时，片段将看到该消息，并执行您想要的任何操作。

注意：可能存在拼写错误。这是从非常大而复杂的代码中删除的。

Answer 3

问题是当你调用execute时，会返回AsyncTask对象，但还没有返回结果。结果在后台计算。结果的类型最终将是一个String（如您所指定的），并将传递给onPostExecute()。

您应该使用AsyncTask，如下所示：

public class ApiAccess extends AsyncTask<List<NameValuePair>, Integer, String> {
    ...
    private String POST(List<NameValuePair>[] nameValuePairs){
    ...
        return response;
    }

    protected void onPreExecute (){
        // this is run on the main (UI) thread, before doInBackground starts
    }

    protected void onPostExecute (String result){
        // this is run on the main (UI) thread, after doInBackground returns
    }

    protected String doInBackground(List<NameValuePair>... nameValuePairs) {
        // run in another, background thread
        return POST(params);
    }
}

请注意，在您的示例中，您不会在doInBackground()中返回结果，您应该这样做。

Answer 4

请阅读AsyncTask。您可以使用onPostExecute方法获得结果。你做不了类似的事情：

String serverResponse = apiObj.execute(nameValuePairs);

因为它是异步的。

如何处理AsyncTask的返回值

4 个答案: