如何在 EventBus 中使用 Call Type

How to use Call Type with EventBus

我正在使用 EventBus 在收到服务器响应时通知 Activity/Fragment。到目前为止一切正常,但是当我在同一个 FragmentActivity 中使用两个网络调用时出现问题。问题是相同的方法 onEvent(String response) 从服务器获取对两个响应的调用。 call 1 的响应不同于 call 2

我想出了一个解决方案 - 我在 NetworkReqest 中添加了 CallType 但我无法通知 activity/fragment 关于网络调用,因为 post() 只需要一个参数.

这里是相关代码-

public class NetworkRequest {
    EventBus eventBus = EventBus.getDefault();

    public void stringParamRequest(String url, final Map<String, String> params,String callType) {
        StringRequest jsonObjRequest = new StringRequest(Request.Method.POST, url,
                new Response.Listener<String>() {
                    @Override
                    public void onResponse(String response) {
                        eventBus.post(response);
                    }
                }, new Response.ErrorListener() {

            @Override
            public void onErrorResponse(VolleyError error) {
                VolleyLog.d("volley", "Error: " + error.getMessage());
                eventBus.post(error);
            }
        }) {

            @Override
            public String getBodyContentType() {
                return "application/x-www-form-urlencoded; charset=UTF-8";
            }

            @Override
            protected Map<String, String> getParams() throws AuthFailureError {
                Map<String, String> param = params;
                return param;
            }

        };
        SkillSchoolApplication.get().addToRequestQueue(jsonObjRequest);
    }

    public void stringRequest(String url, String callType) {
        StringRequest stringRequest = new StringRequest(url, new Response.Listener<String>() {
            @Override
            public void onResponse(String response) {
                eventBus.post(response);
            }
        }, new Response.ErrorListener() {
            @Override
            public void onErrorResponse(VolleyError error) {

            }
        });
        SkillSchoolApplication.get().addToRequestQueue(stringRequest);
    }


}

方法在 fragment/activity 中出现问题,当我从一个请求获得响应后,我触发另一个请求,该请求依赖于第一个请求的响应

@Subscribe(threadMode = ThreadMode.MAIN)
    public void onEvent(String response) {
        Log.d(TAG, response);
        boolean isCourseAvaible = false;
        if (!isCourseAvaible) {
            isCourseAvaible = true;
            List<CoursesDTO> coursesDTOs = AppMgr.coursesMgr(response);
            String[] ids = new String[0];
            String id;
            if (coursesDTOs != null) {
                ids = new String[coursesDTOs.size()];
                for (int i = 0; i < coursesDTOs.size(); i++) {
                    ids[i] = coursesDTOs.get(i).getListId();

                }
            }
            id = TextUtils.join(",", ids);
            Map<String, String> map = new HashMap<>();
            map.put("part", "snippet,contentDetails");
            map.put("playlistId", id);
            map.put("key", AppConstants.YOUTUBE_KEY);
            NetworkRequest networkRequest = new NetworkRequest();
            networkRequest.stringParamRequest("some url", map);
        } else {
            Log.d(TAG, response);
        }

    }

    @Subscribe(threadMode = ThreadMode.MAIN)
    public void onEvent(VolleyError error) {
        Log.d(TAG, error.toString());
        Toast.makeText(getActivity(), "Something went wrong " + error.toString(), Toast.LENGTH_SHORT).show();
    }

如何区分 onEvent() 中的 callType。需要一些指导。非常感谢。

一种选择是将您需要的两条数据包装成一个 class 并将其传递给事件总线。为简洁起见,省略私有成员、getters/setters 和构造函数。

class NetworkResponse() {
   public String callType;
   public String response;
}

收到响应后,分配一个 NetworkResponse 并用响应和调用类型填充它,post 将其填充到事件总线。

@Subscribe(threadMode = ThreadMode.MAIN)
public void onEvent(NetworkResponse networkResponse) {
  if(networkResponse.callType.equals(CALL_1)) {
     // ...
  } else if (networkResponse.callType.equals(CALL_2)) {
     // ...
  }

}

在 onResponse 方法中将字符串响应序列化为 java bean,并将正确的对象发送到视图。 Activity、Fragments 和 Views 无需了解序列化,此外,您的应用程序的性能可以提高,因为您可以修改代码以在后台线程中序列化数据。

public void stringRequest(String url, String callType) {
        StringRequest stringRequest = new StringRequest(url, new Response.Listener<String>() {
            @Override
            public void onResponse(String response) {
                eventBus.post(AppMgr.coursesMgr(response));
            }
        }, new Response.ErrorListener() {
            @Override
            public void onErrorResponse(VolleyError error) {

            }
        });
        SkillSchoolApplication.get().addToRequestQueue(stringRequest);
    }

那么您的第一个活动订阅将如下所示:

@Subscribe(threadMode = ThreadMode.MAIN)
    public void onEvent(List<CoursesDTO> coursesDTOs) {
        Log.d(TAG, response);
        boolean isCourseAvaible = false;
        if (!isCourseAvaible) {
            isCourseAvaible = true;
            String[] ids = new String[0];
            String id;
            if (coursesDTOs != null) {
                ids = new String[coursesDTOs.size()];
                for (int i = 0; i < coursesDTOs.size(); i++) {
                    ids[i] = coursesDTOs.get(i).getListId();
                }
            }
            id = TextUtils.join(",", ids);
            Map<String, String> map = new HashMap<>();
            map.put("part", "snippet,contentDetails");
            map.put("playlistId", id);
            map.put("key", AppConstants.YOUTUBE_KEY);
            NetworkRequest networkRequest = new NetworkRequest();
            networkRequest.stringParamRequest("some url", map);
        } else {
            Log.d(TAG, response);
        }
    }

并且您可以有第二个不同的字符串订阅。不过反正你要打第二个,所以最好在第一个正确答案后直接执行。

public class NetworkRequest {
    EventBus eventBus = EventBus.getDefault();

    public void stringParamRequest(String url, final Map<String, String> params,String callType) {
        StringRequest jsonObjRequest = new StringRequest(Request.Method.POST, url,
                new Response.Listener<String>() {
                    @Override
                    public void onResponse(String response) {
                        eventBus.post(response);
                    }
                }, new Response.ErrorListener() {

            @Override
            public void onErrorResponse(VolleyError error) {
                VolleyLog.d("volley", "Error: " + error.getMessage());
                eventBus.post(error);
            }
        }) {

            @Override
            public String getBodyContentType() {
                return "application/x-www-form-urlencoded; charset=UTF-8";
            }

            @Override
            protected Map<String, String> getParams() throws AuthFailureError {
                Map<String, String> param = params;
                return param;
            }

        };
        SkillSchoolApplication.get().addToRequestQueue(jsonObjRequest);
    }

    public void stringRequest(String url, String callType) {
        StringRequest stringRequest = new StringRequest(url, new Response.Listener<String>() {
            @Override
            public void onResponse(String response) {  
              List<CoursesDTO> coursesDTOs = AppMgr.coursesMgr(response);
              String[] ids = new String[0];
              String id;
              if (coursesDTOs != null) {
                ids = new String[coursesDTOs.size()];
                for (int i = 0; i < coursesDTOs.size(); i++) {
                    ids[i] = coursesDTOs.get(i).getListId();

                }
              }
              id = TextUtils.join(",", ids);
              Map<String, String> map = new HashMap<>();
              map.put("part", "snippet,contentDetails");
              map.put("playlistId", id);
              map.put("key", AppConstants.YOUTUBE_KEY);
              NetworkRequest networkRequest = new NetworkRequest();
              networkRequest.stringParamRequest("some url", map);                   
            }
        }, new Response.ErrorListener() {
            @Override
            public void onErrorResponse(VolleyError error) {

            }
        });
        SkillSchoolApplication.get().addToRequestQueue(stringRequest);
    }
}

最后,这两行

  boolean isCourseAvaible = false;
        if (!isCourseAvaible) {

多余的,就像没有条件一样。