我正在使用 Retrofit2 和 RxJava2CallAdapterFactory。
我使用的 API 始终返回 200 状态代码，对于成功和响应 JSON 字符串，JSON 结构完全不同。由于状态代码始终为 200，因此始终调用 onResponse() 方法。因此，我无法在错误情况下从 JSON 中提取错误消息。
解决方案一:
我使用 ScalarsConverterFactory 获取响应字符串并手动使用 Gson 来解析响应。
How to get response as String using retrofit without using GSON or any other library in android
此解决方案的问题:我计划使用 RxJava2CallAdapterFactory，因为 retrofit 方法应返回 DataModel 类。
我需要找到这个问题的最佳解决方案，这样我就可以继续从 Retrofit 方法返回数据模型类，并以某种方式从响应中识别错误条件(识别响应 JSON 与数据模型不匹配)，然后将错误 JSON 解析为数据模型。
retrofit 客户端
public static Retrofit getClient(String url) {
if (apiClient == null) {
HttpLoggingInterceptor interceptor = new HttpLoggingInterceptor();
interceptor.setLevel(HttpLoggingInterceptor.Level.BODY);
OkHttpClient httpClient = new OkHttpClient.Builder().addInterceptor(interceptor).build();
apiClient = new Retrofit.Builder()
.baseUrl(url)
/*addCallAdapterFactory for RX Recyclerviews*/
.addCallAdapterFactory(RxJava2CallAdapterFactory.create())
/* add ScalarsConverterFactory to get json string as response */
// .addConverterFactory(ScalarsConverterFactory.create())
.addConverterFactory(GsonConverterFactory.create())
// .addConverterFactory(GsonConverterFactory.create(gson))
.client(httpClient)
.build();
}
return apiClient;
}
方法
public static void getLoginAPIResponse(String username, String password, String sourceId, String uuid, final HttpCallback httpCallback) {
baseUrl = AppPreference.getParam(UiUtils.getContext(), SPConstants.BASE_URL, "").toString();
ApiInterface apiService =
ApiClient.getClient(baseUrl).create(ApiInterface.class);
Call<LoginBean> call = apiService.getLoginResponse(queryParams);
call.enqueue(new Callback<LoginBean>() {
@Override
public void onResponse(Call<LoginBean> call, Response<LoginBean> response) {
if (response.body().isObjectNull()) {
httpCallback.resultCallback(APIConstants.API_LOGIN, HttpCallback.REQUEST_TYPE_GET,
HttpCallback.RETURN_TYPE_FAILURE, 0, null);
return;
}
httpCallback.resultCallback(APIConstants.API_LOGIN, HttpCallback.REQUEST_TYPE_GET,
HttpCallback.RETURN_TYPE_SUCCESS, response.code(), response.body());
}
@Override
public void onFailure(Call<LoginBean> call, Throwable t) {
// Log error here since request failed
httpCallback.resultCallback(APIConstants.API_APP_VERIFICATION, HttpCallback.REQUEST_TYPE_GET,
HttpCallback.RETURN_TYPE_FAILURE, 0, t);
t.printStackTrace();
}
});
}
界面
@GET("App/login")
Call<LoginBean> getLoginResponse(@QueryMap Map<String, String> queryMap);
附言: API 目前无法更改，因为一些其他应用程序正在使用它。
Gson 解析器不返回 null 对象实例让我理解存在 json 结构和数据模型不匹配。
RestAdapter 在 Retrofit 2 中被弃用
我正在寻找解决此问题的最佳方法，最好避免手动 json 解析并充分利用改造和 RX 适配器。
编辑
因此响应代码为 200
response.isSuccessful() == trueresponse.body() != null也是如此，因为如果 JSON 结构不匹配，Gson 永远不会创建 null 实例或抛出任何异常response.errorBody() == null始终作为从服务器作为输入流发送的响应。if (response.isSuccessful() && response.body() != null) { //control always here as status code 200 for error condition also }else if(response.errorBody()!=null){ //control never reaches here }
编辑 2
解决方案
解决方案基于anstaendig答案
- 我创建了一个基础泛型类来进一步回答这个问题。
- 因为我有多个 api 和数据模型，所以我必须为每个创建反序列化器
基础 API BEAN
public class BaseApiBean<T> {
@Nullable
private T responseBean;
@Nullable
private ErrorBean errorBean;
public BaseApiBean(T responseBean, ErrorBean errorBean) {
this.responseBean = responseBean;
this.errorBean = errorBean;
}
public T getResponseBean() {
return responseBean;
}
public void setResponseBean(T responseBean) {
this.responseBean = responseBean;
}
public ErrorBean getErrorBean() {
return errorBean;
}
public void setErrorBean(ErrorBean errorBean) {
this.errorBean = errorBean;
}
}
基础解串器
public abstract class BaseDeserializer implements JsonDeserializer<BaseApiBean> {
@Override
public BaseApiBean deserialize(JsonElement json, Type typeOfT, JsonDeserializationContext context)
throws JsonParseException {
// Get JsonObject
final JsonObject jsonObject = json.getAsJsonObject();
if (jsonObject.has("result")) {
/* {"result":"404"}*/
ErrorBean errorMessage = new Gson().fromJson(jsonObject, ErrorBean.class);
return getResponseBean(errorMessage);
} else {
return getResponseBean(jsonObject);
}
}
public abstract BaseApiBean getResponseBean(ErrorBean errorBean);
public abstract BaseApiBean getResponseBean(JsonObject jsonObject);
}
为每个 API 定制反序列化器
public class LoginDeserializer extends BaseDeserializer {
@Override
public BaseApiBean getResponseBean(ErrorBean errorBean) {
return new LoginResponse(null, errorBean);
}
@Override
public BaseApiBean getResponseBean(JsonObject jsonObject) {
LoginBean loginBean = (new Gson().fromJson(jsonObject, LoginBean.class));
return new LoginResponse(loginBean, null);
}
}
自定义响应 BEAN
public class LoginResponse extends BaseApiBean<LoginBean> {
public LoginResponse(LoginBean responseBean, ErrorBean errorBean) {
super(responseBean, errorBean);
}
}
客户
public class ApiClient {
private static Retrofit apiClient = null;
private static Retrofit apiClientForFeedBack = null;
private static LoginDeserializer loginDeserializer = new LoginDeserializer();
private static AppVerificationDeserializer appVerificationDeserializer = new AppVerificationDeserializer();
public static Retrofit getClient(String url) {
if (apiClient == null) {
GsonBuilder gsonBuilder=new GsonBuilder();
gsonBuilder.registerTypeAdapter(LoginResponse.class,
loginDeserializer);
gsonBuilder.registerTypeAdapter(AppVerificationResponse.class,
appVerificationDeserializer);
Gson gson= gsonBuilder.create();
HttpLoggingInterceptor interceptor = new HttpLoggingInterceptor();
interceptor.setLevel(HttpLoggingInterceptor.Level.BODY);
OkHttpClient httpClient = new OkHttpClient.Builder().addInterceptor(interceptor)
.retryOnConnectionFailure(true)
.connectTimeout(15, TimeUnit.SECONDS)
.build();
apiClient = new Retrofit.Builder()
.baseUrl(url)
/*addCallAdapterFactory for RX Recyclerviews*/
.addCallAdapterFactory(RxJava2CallAdapterFactory.create())
/* add ScalarsConverterFactory to get json string as response */
// .addConverterFactory(ScalarsConverterFactory.create())
// .addConverterFactory(GsonConverterFactory.create())
.addConverterFactory(GsonConverterFactory.create(gson))
.client(httpClient)
.build();
}
return apiClient;
}
处理响应
public static void getLoginAPIResponse(String username, String password, String sourceId, String uuid, final HttpCallback httpCallback) {
baseUrl = AppPreference.getParam(getContext(), SPConstants.MT4_BASE_URL, "").toString();
ApiInterface apiService =
ApiClient.getClient(baseUrl).create(ApiInterface.class);
HashMap<String, String> queryParams = new HashMap<>();
queryParams.put(APIConstants.KEY_EMAIL, sourceId + username.toLowerCase());
queryParams.put(APIConstants.KEY_PASSWORD, Utils.encodePwd(password));
Call<LoginResponse> call = apiService.getLoginResponse(queryParams);
call.enqueue(new Callback<LoginResponse>() {
@Override
public void onResponse(Call<LoginResponse> call, Response<LoginResponse> response) {
if (response.body().getResponseBean()==null) {
httpCallback.resultCallback(APIConstants.API_LOGIN, HttpCallback.REQUEST_TYPE_GET,
HttpCallback.RETURN_TYPE_FAILURE, 0, response.body().getErrorBean());
return;
}
httpCallback.resultCallback(APIConstants.API_LOGIN, HttpCallback.REQUEST_TYPE_GET,
HttpCallback.RETURN_TYPE_SUCCESS, response.code(), response.body().getResponseBean());
}
@Override
public void onFailure(Call<LoginResponse> call, Throwable t) {
// Log error here since request failed
httpCallback.resultCallback(APIConstants.API_APP_VERIFICATION, HttpCallback.REQUEST_TYPE_GET,
HttpCallback.RETURN_TYPE_FAILURE, 0, t);
t.printStackTrace();
}
});
}
最佳答案
因此您有来自同一端点的两个不同的成功(状态代码 200)响应。一个是实际的数据模型，一个是错误的(都是像这样的 json 结构？:
有效的 LoginBean 响应:
{
"id": 1234,
"something": "something"
}
错误响应
{
"error": "error message"
}
您可以做的是拥有一个包装这两种情况并使用自定义反序列化器的实体。
class LoginBeanResponse {
@Nullable private final LoginBean loginBean;
@Nullable private final ErrorMessage errorMessage;
LoginBeanResponse(@Nullable LoginBean loginBean, @Nullable ErrorMessage errorMessage) {
this.loginBean = loginBean;
this.errorMessage = errorMessage;
}
// Add getters and whatever you need
}
错误的包装器:
class ErrorMessage {
String errorMessage;
// And whatever else you need
// ...
}
然后你需要一个JsonDeserializer :
public class LoginBeanResponseDeserializer implements JsonDeserializer<LoginBeanResponse> {
@Override
public LoginBeanResponse deserialize(JsonElement json, Type typeOfT, JsonDeserializationContext context) throws JsonParseException {
// Based on the structure you check if the data is valid or not
// Example for the above defined structures:
// Get JsonObject
final JsonObject jsonObject = json.getAsJsonObject();
if (jsonObject.has("error") {
ErrorMessage errorMessage = new Gson().fromJson(jsonObject, ErrorMessage.class);
return new LoginBeanResponse(null, errorMessage)
} else {
LoginBean loginBean = new Gson().fromJson(jsonObject, LoginBean.class):
return new LoginBeanResponse(loginBean, null);
}
}
}
然后将这个反序列化器添加到 GsonConverterFactory :
GsonBuilder gsonBuilder = new GsonBuilder().registerTypeAdapter(LoginBeanResponse.class, new LoginBeanResponseDeserializer()).create():
apiClient = new Retrofit.Builder()
.baseUrl(url)
.addCallAdapterFactory(RxJava2CallAdapterFactory.create())
.addConverterFactory(GsonConverterFactory.create(gsonBuilder))
.client(httpClient)
.build();
这是我能想到的使这项工作成功的唯一方法。但正如已经提到的，这种 API 设计是错误的，因为状态代码的存在是有原因的。我仍然希望这会有所帮助。
编辑:然后您可以在调用该 Retrofit 的类中执行的操作(如果您已经使用 RxJava 从 Call<LoginBeanResponse> 转换为 Single<LoginBeanResponse>)实际上是返回一个正确的错误。像这样的东西:
Single<LoginBean> getLoginResponse(Map<String, String> queryMap) {
restApi.getLoginResponse(queryMap)
.map(loginBeanResponse -> { if(loginBeanResponse.isError()) {
Single.error(new Throwable(loginBeanResponse.getError().getErrorMessage()))
} else {
Single.just(loginBeanReponse.getLoginBean())
}})
}
关于java - Retrofit2 状态码为 200 但 json 结构不同于数据模型类时的处理条件，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/42802951/