19

我正在使用AsyncTask具有以下签名的类：

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

但在这里我得到这个错误：

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

为什么当我String在类扩展行中指定为第三个参数时？

4

4 回答 4

34

您可以通过在返回的 AsyncTask 上调用 AsyhncTask 的 get() 方法来获取结果，但它会在等待获取结果时将其从异步任务变为同步任务。

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

由于您将 AsyncTask 放在一个单独的类中，因此您可以创建一个接口类并在 AsyncTask 中声明它，并将您的新接口类实现为您希望从中访问结果的类中的委托。一个很好的指南在这里：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);
}

（API访问）

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");
        }
    } 
}

（你的主类，它实现了 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.
}
于 2013-10-27T22:17:42.413 回答
3

我建议实现一个处理程序回调。您可以将片段（或活动）的处理程序传递给 AsyncTask，AsyncTask 将在完成时调用它。AsyncTask 也可以传回任意对象。

这是一个示例 AsyncTask，我在它自己的文件中（不是子类）：

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);
    }
}

此示例假定您的 AsyncTask 需要一块糖果和一根冰棒。然后它将向您的片段返回一个甜点。

您可以使用以下代码在片段中的一行中构造和运行 AsyncTask：

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

但当然，您首先需要设置片段的处理程序 (myFragmentHandler)。为此，您的片段（或活动）应如下所示（注意“实现 Handler.Callback”）：

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。调用片段将在 AsyncTask 处理期间继续执行。然后，当 AsyncTask 完成时，它会向 Fragment 发送一条消息，表示它已完成，并随消息传递一个对象。此时，fragment 将看到消息，并为所欲为。

注意：可能有错别字。这是从一个非常大而复杂的代码中剪下来的。

于 2014-10-01T10:47:47.607 回答
0

问题是当你调用 execute 时，AsyncTask 对象被返回，但还没有结果。结果在后台计算。结果的类型最终将是一个字符串（如您指定的那样），并将传递给onPostExecute().

你应该使用AsyncTask如下：

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，您没有返回结果doInBackground()，这是您应该的。

于 2013-10-27T22:46:44.703 回答
-1

请阅读AsyncTask。您可以在onPostExecute方法上获得结果。你不能做这样的事情：

String serverResponse = apiObj.execute(nameValuePairs);

因为它是异步的。

于 2013-10-27T22:11:10.527 回答