【发布时间】:2015-12-28 18:04:57
【问题描述】:
我在这里第一次使用 Retrofit。
我想将我的 List 从 Callback 放到 UsersData 类中。这是不可能的。但是,如果我从用户数据中删除所有内容并将配置文件中的内容放入用户数据中,那么它就可以工作。但这并不能满足我的需求。我需要能够将 List 从 Callback 放到 UsersData 类中。
谢谢你的好处
在我的片段中
App.getRestClient().getAttendanceService().getUsers(48, new Callback<List<UsersData>>() {
@Override
public void success(List<UsersData> usersDao, Response response) {
String ble = usersDao.get(0).getResults().get(0).getFirstName();
Toast.makeText(getActivity(),ble, Toast.LENGTH_SHORT).show();
}
@Override
public void failure(RetrofitError error) {
}
});
应用程序
public class App extends Application {
private static RestClient restClient;
public static App instance = null;
public static Context getInstance() {
if (null == instance) {
instance = new App();
}
return instance;
}
@Override
public void onCreate(){
super.onCreate();
restClient = new RestClient();
}
public static RestClient getRestClient(){
return restClient;
}
}
还有我的客户
public class RestClient {
private static final String BASE_URL = "www.Link_to_json.com" ;
private AttendanceService attendanceService;
public RestClient()
{
Gson gson = new GsonBuilder()
.setDateFormat("yyyy'-'MM'-'dd'T'HH':'mm':'ss'.'SSS'Z'")
.create();
RestAdapter restAdapter = new RestAdapter.Builder()
//.setLogLevel(RestAdapter.LogLevel.FULL)
.setEndpoint(BASE_URL)
//.setClient(new OkClient(new OkHttpClient()))
//.setConverter(new GsonConverter(gson))
.build();
attendanceService = restAdapter.create(AttendanceService.class);
}
public AttendanceService getAttendanceService()
{
return attendanceService;
}
}
我的界面
public interface AttendanceService {
@GET("/GetUsers")
void getUsers(@Query("companyId") int i, Callback<List<UsersData>> u );
}
和用户数据
public class UsersData {
private List<Profile> results;
public List<Profile> getResults() {
return results;
}
}
个人资料数据类:
public String firstName;
public String lastname;
public int userId;
public String userNameId;
...
json 示例:
[
{
"AttendanceDate":null,
"AttendanceStatus":1,
"AttendanceStatusDescription":null,
"CompanyId":48,
"Email":"",
"FirstName":"Sindri",
"Gender":1,
"Gsm":"",
"Id":259,
"LastName":"yeh",
"MiddleName":"",
"Role":0,"UserId":"corp\\marg"
},{
"AttendanceDate":null,
"AttendanceStatus":1,
"AttendanceStatusDescription":null,
"CompanyId":48,
"Email":"",
"FirstName":"David",
"Gender":1,
"Gsm":"",
"Id":165,
"LastName":"Guðmundsson",
"MiddleName":"",
"Role":0,"UserId":"corp\\marg"
}
]
【问题讨论】:
-
尝试给json数组一个key或name,并在model中分别处理