Question

我正在尝试使用OkHTTP库。调用服务器并获得成功响应时。我需要更新UI。

使用enqueue（）进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，希望这有帮助！

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

If your code doesn't update the UI, I would suggest you to specify the thread, since UI is on its own thread: 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

Try this:

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

Answer 4

为了拥有干净的代码，我建议你不要将整个代码放在runnable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
    }

}

OkHTTP从enqueue回调更新UI

4 个答案: