如何将改造响应体转换为 JsonObject答案

【问题标题】：How to convert retrofit response body into JsonObject如何将改造响应体转换为 JsonObject
【发布时间】：2016-09-28 05:38:22
【问题描述】：

public abstract class RetrofitBaseCallBack<T> implements Callback<T> {
    @Override
    public void onResponse(Call<T> call, Response<T> response) {
        int responseCode = response.code();
        if (responseCode == Constants.ServerResponse.Code.STATUS_OK
                || responseCode == Constants.ServerResponse.Code.STATUS_CREATED_SUCCESS
                || responseCode == Constants.ServerResponse.Code.STATUS_IN_PROGRESS
                || responseCode == Constants.ServerResponse.Code.STATUS_NO_CONTENT
                || responseCode == Constants.ServerResponse.Code.STATUS_ACCEPTED) {
            // i want to navigate to onResponseSuccess when "success":true
            onResponseSuccess(call, response);
        } else {
           //i want to navigate to onResponseFailure when "success":false
            onErrorResponse(call, response);
        }
    }
    }

    @Override
    public void onFailure(Call<T> call, Throwable t) {
        onRetrofitBaseCallBackFailure(call, t);
    }

    public abstract void onResponseSuccess(Call<T> call, Response<T> response);

    public abstract void onResponseFailure(Call<T> call, Response response, String message);

    public abstract void onRetrofitBaseCallBackFailure(Call<T> call, Throwable t);
}

我通过扩展改造回调来上课。我想将响应解析为字符串。在我的例子中，每个响应都会获得这种类型的 JSON 对象。在成功响应的情况下：

{
  "success": true,
  "abc": {
    "a": 0.15,
    "b": "Better Exchange Rate",
  }
}

如果失败响应

{
      "success": false,
      "abc": {
        "a": 0.15,
        "b": "Better Exchange Rate",
      }
    }

如何将响应正文转换为 JsonObject 以便我可以检查字段 "success": 并导航相应的方法，即 onResponseSuccess(Call call, Response response) 以防万一成功时，onResponseFailure(Call call, Response response, String message) 失败时

【问题讨论】：

标签： android gson retrofit2

【解决方案1】：

如何在改造中从 response.body() 获取 JSONObject：

try {

    JSONObject jsonObj = new JSONObject(response.body().toString());

} catch (JSONException e) {

    e.printStackTrace();
    Log.d("TAG", "No valid json");

}

【讨论】：