Question

我知道这是一个常见问题，但我都做了，但仍未解决。在我的 MainActivity 中，我收到了来自 ServerService.java 的呼叫，如下所示：

String randomNumber = serverService.contactServer();

在 ServerService.java 中， contactServer（）将调用包含.enqueue的方法：

  public String contactServer() {     
    return requestServerService();
}

requestServerService（）包含以下代码：

 public String requestServerService() {

    Call<RequestAttributes> call = new RetrofitConfig().getServiceRequester().requestRandomNumber();

    call.enqueue(new Callback<RequestAttributes>() {
        @Override
        public void onResponse(Call<RequestAttributes> call, Response<RequestAttributes> response) {
            if (!response.isSuccessful()) {
                Log.i("Err", "Err: " + response.code());
            } else {
                RequestAttributes requestAttributes = response.body();
                returnedValue = requestAttributes.getRandomNumber();
                Log.d("jsonAnswer", "O numero aleatorio é: " + returnedValue);
            }
        }

        @Override
        public void onFailure(Call<RequestAttributes> call, Throwable t) {
            Log.e("Fail", "Failed: " + t.getMessage());
        }
    }); return returnedValue;

错误是 returnedValue 返回空值。我尝试了调试，但即使没有达到onReponse。我知道问题一定是因为.enqueue是异步的，但是如何解决此问题并将请求返回给 mainActivity ？

改造的配置：

  public RetrofitConfig() {

    this.retrofit = new Retrofit.Builder()
                    .baseUrl("localhost:3000/")
                    .addConverterFactory(GsonConverterFactory.create())
                    .build();

}

public ServiceRequester getServiceRequester() {
    return this.retrofit.create(ServiceRequester.class);
}

POJO ：

public class RequestAttributes {


@SerializedName("randomNumber")
private String randomNumber;

public String getRandomNumber() {
    return randomNumber;
}

public void setRandomNumber(String randomNumber) {
    this.randomNumber = randomNumber;
}

@Override
public String toString() {
    return "RequestAttributes{" +     
            ", randomNumber='" + randomNumber + '\'' +
            '}';
}

}

和请求：

 @GET("api/requestRandomNumber")
 Call<RequestAttributes> requestRandomNumber();

JSON答案（如果我通过浏览器请求）

{"randomNumber":"u845gq"}

Answer 1

您可以将回调从MainActivity传递到contactServer()方法

        serverService.contactServer(new Callback<RequestAttributes>() {
        @Override
        public void onResponse(Call<RequestAttributes> call, Response<RequestAttributes> response) {
            if (!response.isSuccessful()) {
                Log.i("Err", "Err: " + response.code());
            } else {
                RequestAttributes requestAttributes = response.body();
                String returnedValue = requestAttributes.getRandomNumber();
                // Do what you want here with returnedValue. You are in the activity thread(MainThread or UIThread) for example someTextView.setText(returnedValue);
                Log.d("jsonAnswer", "O numero aleatorio é: " + returnedValue);
            }
        }

        @Override
        public void onFailure(Call<RequestAttributes> call, Throwable t) {
            Log.e("Fail", "Failed: " + t.getMessage());
        }
    });

然后将其设为void方法，并将回调传递给requestServerService()方法

public void contactServer(Callback<RequestAttributes> callback) {
    requestServerService(callback);
}

然后实现如下的requestServerService()方法：

public void requestServerService(Callback<RequestAttributes> callback) {

    Call<RequestAttributes> call = new RetrofitConfig().getServiceRequester().requestRandomNumber();

    call.enqueue(callback);
}

翻新呼叫队列未达到onResponse或onFailure

1 个答案: