Android的改造2自定义JSON请求

我在我使用改装2.开发一个应用程序我的要求JSON是如下Android的改造2自定义JSON请求

{ 
    "FirstName":"ABC", 
    "LastName":"XYZ", 
    "Email":"[email protected]", 
    "Password":123456", 
    "PhoneNo":"1234567890", 
    "Country":"1", 
    "State":"1", 
    "City":"1", 
    "Town":"YYYYYY", 
    "Details":[{ 
     "Education":"BE", 
     "Sports":"Cricket", 
     "Hobby":"TV" 
    }] 
}

我的API服务调用如下

public interface APIService { 
    @POST("/signup") 
    Call<SignUpResponseModel> signUp(@Body SignUpRequestParent body); 
}

我API Util类如下

public class ApiUtils { 
    private ApiUtils() {} 
    public static final String BASE_URL = "http://URL/"; 

    public static APIService getAPIService() { 
     return RetrofitClient.getClient(BASE_URL).create(APIService.class); 
    } 
}

我的改造客户端类如下：

public class RetrofitClient { 
    private static Retrofit retrofit = null; 
    public static Retrofit getClient(String baseUrl) { 
     if (retrofit==null) { 
      retrofit = new Retrofit.Builder() 
        .client(getHeader()) 
        .baseUrl(baseUrl) 
        .addConverterFactory(GsonConverterFactory.create()) 
        .build(); 
     } 
     return retrofit; 
    } 

    public static OkHttpClient getHeader() { 
     HttpLoggingInterceptor interceptor = new HttpLoggingInterceptor(); 
     interceptor.setLevel(HttpLoggingInterceptor.Level.BODY); 
     OkHttpClient okClient = new OkHttpClient.Builder() 
       .addInterceptor(interceptor) 
       .addNetworkInterceptor(new Interceptor() { 
        @Override 
        public Response intercept(Interceptor.Chain chain) throws IOException { 
         Request request = null; 
         Request original = chain.request(); 
         // Request customization add request headers 
         Request.Builder requestBuilder = original.newBuilder() 
          .addHeader("Authorization", "auth") 
         request = requestBuilder.build(); 
         return chain.proceed(request); 
        } 
       }).build(); 
     return okClient; 
    } 
}

我的要求模型类如下：

public class SignUpRequestParent { 
    final String FirstName; 
    final String LastName; 
    final String Email; 
    final String Password; 
    final String PhoneNo; 
    final String Country; 
    final String State; 
    final String City; 
    final String Town; 
    final List<SignUpRequestChild> Details; 
    public SignUpRequestParent(String FirstName, String LastName, String Email, String Password, String PhoneNo, String Country, String State, String City, String Town, List<SignUpRequestChild> Details) { 
     this.FirstName = FirstName; 
     this.LastName = LastName; 
     this.Email = Email; 
     this.Password = Password; 
     this.PhoneNo = PhoneNo; 
     this.Country = Country; 
     this.State = State; 
     this.City = City; 
     this.Town = Town; 
     this.Details = Details; 
    } 
} 

public class SignUpRequestChild { 
    final String Education; 
    final String Sports; 
    final String Hobby; 

    public SignUpRequestChild(String Education, String Sports, String Hobby) { 
     this.Education = Education; 
     this.Sports = Sports; 
     this.Hobby = Hobby; 
    } 
}

下面就是我得到错误 Web服务的调用代码。

mAPIService.signUp(new SignUpRequestParent(name, surName, email, password, mobileNumber, country, state, city, Town, new SignUpRequestChild(Education,Sports,Hobby))).enqueue(new Callback<SignUpResponseModel>() { 
    @Override 
    public void onResponse(Call<SignUpResponseModel> call, Response<SignUpResponseModel> response) { 
     if(response.isSuccessful()) {                   
      Log.e("SignUpResponse", response.body().toString()); 
     } 
    } 
    @Override 
    public void onFailure(Call<SignUpResponseModel> call, Throwable t) { 
     Log.e("RetroFit", ""+t); 
    } 
});

调用代码被赋予误差的第一行。因为我不知道如何从类创建嵌套的JSON数组。

请指点我该怎么做。而我错了。

预先感谢您。

来源

2017-09-15 Manoj Fegde

可否请你用错误添加堆栈跟踪？ –

也SignUpResponseModel定义 –

@cgomezmendez - 我无法打电话。编码时的错误是：将第11个参数投射到Java.util.List。或者从List 更改为SignUpRequestChild的方法'SignUpRequestParent'的第11个参数。 –

拨打因此，我认为你的问题是你这样做

mAPIService.signUp(new SignUpRequestParent(name, surName, 
email, password, mobileNumber, country, state, city, 
    Town, new SignUpRequestChild(Education,Sports,Hobby))).enqueue(new Callback<SignUpResponseModel>() { 
    @Override 
    public void onResponse(Call<SignUpResponseModel> call, 
    Response<SignUpResponseModel> response) { 
     if(response.isSuccessful()) {                   
      Log.e("SignUpResponse", response.body().toString()); 
     } 
    } 
    @Override 
    public void onFailure(Call<SignUpResponseModel> call, Throwable t) { 
     Log.e("RetroFit", ""+t); 
    } 
});

但你

new SignUpRequestChild(Education,Sports,Hobby))

它有望成为一个列表，按您的构造函数，所以你确实应该这样做

List childs = new ArrayList<SignUpRequestChild>(); 
childs.add(new SignUpRequestChild(Education,Sports,Hobby))); 
mAPIService.signUp(new SignUpRequestParent(name, surName, 
email, password, mobileNumber, country, state, city, 
    Town, childs).enqueue(new Callback<SignUpResponseModel>() { 
    @Override 
    public void onResponse(Call<SignUpResponseModel> call, 
    Response<SignUpResponseModel> response) { 
     if(response.isSuccessful()) {                   
      Log.e("SignUpResponse", response.body().toString()); 
     } 
    } 
    @Override 
    public void onFailure(Call<SignUpResponseModel> call, Throwable t) { 
     Log.e("RetroFit", ""+t); 
    } 
});

来源

2017-09-15 11:56:30

感谢您的回复。我认为你的答案正在工作，但现在我得到了java.net。SocketTimeoutException。 –

你可以试试https://stackoverflow.com/questions/39219094/sockettimeoutexception-in-retrofit –

感谢兄弟。它的工作现在。 –

您应该提供带有错误的堆栈跟踪。通过浏览代码我可以看到的一件事是SignUpRequestParent构造函数接受SignUpRequestChild对象的列表，而您在进行api调用时传递一个新的SignUpRequestChild对象。

来源

2017-09-15 11:22:06

我想你调用了错误的方式，试试这个

在RetrofitClient类中添加这

public static APIService getService() { 
    return getClient().create(APIService.class); 
    }

然后将它从任何地方

RetrofitClient.getService.someMethodHere().enqueue ...

来源

2017-09-15 11:25:37

Android的改造2自定义JSON请求

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