通过排队回调OkHTTP更新UI

Question

我正在尝试使用OkHTTP库。当对服务器进行呼叫并获得成功回复时。我需要更新UI。

使用enqueue（）执行异步调用时，如何做到这一点？

client.newCall(request).enqueue(new Callback() {
    @Override
    public void onFailure(Request request, IOException e) {
    }
    @Override
    public void onResponse(Response response) throws IOException {
        if (response.isSuccessful()) {
            // NEED TO UPDATE UI HERE
        } 
    }
});

Answer 1

您可以参考以下示例代码，希望这能有所帮助！

public class MainActivity extends AppCompatActivity {
    private static final String LOG_TAG = "OkHttp";
    private TextView mTextView;
    private Handler mHandler;
    private String mMessage;
    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_main);
        mTextView = (TextView) findViewById(R.id.textView);
        mHandler = new Handler(Looper.getMainLooper());
        OkHttpClient client = new OkHttpClient();
        // GET request
        Request request = new Request.Builder()
                .url("http://...")
                .build();
        client.newCall(request).enqueue(new Callback() {
            @Override
            public void onFailure(Request request, IOException e) {
                mMessage = e.toString();
                Log.e(LOG_TAG, mMessage); // no need inside run()
                mHandler.post(new Runnable() {
                    @Override
                    public void run() {
                        mTextView.setText(mMessage); // must be inside run()
                    }
                });
            }
            @Override
            public void onResponse(Response response) throws IOException {
                mMessage = response.toString();
                Log.i(LOG_TAG, mMessage); // no need inside run()
                mHandler.post(new Runnable() {
                    @Override
                    public void run() {
                        mTextView.setText(mMessage); // must be inside run()
                    }
                });
            }
        });
    }
}

Answer 2

如果您的代码没有更新UI，我建议您指定线程，因为UI在自己的线程上：

client.newCall(request).enqueue(new Callback() {
    @Override
    public void onFailure(Request request, IOException e) {
    }
    @Override
    public void onResponse(Response response) throws IOException {
        if (response.isSuccessful()) {
            runOnUiThread(new Runnable() {
                @Override
                public void run() {
                    //TODO: update your UI
                }
            });
        } 
    }
});

Answer 3

试试这个：

Handler mainHandler = new Handler(Looper.getMainLooper());
mainHandler.post(new Runnable() {
    @Override
    public void run() {
        // code to interact with UI
    }
});

Answer 4

为了拥有一个干净的代码，我建议您不要将整个代码放在可运行程序中。

一个简单的例子：

public class MainActivity extends AppCompatActivity {
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_main);
        OkHttpClient client = new OkHttpClient();
        Request request = new Request.Builder()
            .url("http://...").build();
        client.newCall(request).enqueue(new Callback() {

            @Override
            public void onFailure(Request request, IOException e) {
                e.printStackTrace();
            }

            @Override
            public void onResponse(Response response) throws IOException {
                final String body = response.body().string();
                runOnUiThread(new Runnable() {
                    @Override
                    public void run() {
                        updateUI(body);
                    }
                });
            }

        });
    }
    private void updateUI(responseBody){
        //TODO: update your UI
    }
}