Question

问题：

  Spring在mongodb中创建了两个集合，分别是超类雇员姓名和基类经理和主管。   我想在春季之前阻止在mongo db中为超类员工创建集合

我正在使用Spring-boot JPA mongodb maven插件。
  我试图仅使用一个通用存储库和相同的超级实体类在mongo collection Manager和超级用户上执行插入操作。

manager
{
id:"abc",
name:"Deva",
age:26
}

supervisor
{
id:"abc",
name:"Raj",
age:25
}

这是我的雇员班，是超级班，将由我的经理和主管班扩展

public @Data class Employee
{
  private String id;
  private String name;
  private int age;
}

Manager类绑定到集合管理器

@JsonInclude(JsonInclude.Include.NON_NULL)
@Document(collection="manager")
public @Data class Manager extends Employee
{

}

主管类绑定到集合主管

@JsonInclude(JsonInclude.Include.NON_NULL)
@Document(collection="supervisor")
public @Data class Supervisor extends Employee
{

}

使用一种save（）方法将通用接口实现以将文档保存到集合中

public interface EmployeeService<T> 
{
    public T save(T employee);
}

@Service
public class EmployeeServiceImpl<T> implements EmployeeService<T> 
{
    @Autowired
    EmployeeRepository employeeRepository;

    @Override
    public T save(T obj) {
        return (T) employeeRepository.save(obj);
    }
}

我已经创建了一个存储库EmployeeRepository，它是通用的，可用于经理和主管实体集合文档。

public interface EmployeeRepository<T extends Employee> extends MongoRepository<T, String> {
{

}

这是我的组件，它将调用以保存经理或主管mongo集合中的雇员。

@Component
public class EmployeeComponent<T> 
{

    @Autowired
    EmployeeService employeeService;

    private boolean updateEmployee(Employee employee) throws Exception 
    {

        Employee employees= employeeService.save(employee);

        if (employees.isEmpty())
        {
            System.out.println("Failed to save employees.");
            return "Failed to save employees.";
        }
        else
        {
            System.out.println("Success.");
            return "Success.";
        }
    }
}

这是我的控制器调用保存功能并根据条件传递的员工或主管对象

@RestController
@RequestMapping("test")
public class TestController {

@Autowired
EmployeeComponent<?> employeeComponent;
@Autowired
ObjectMapper mapper;

@RequestMapping(value="save/employee", method=RequestMethod.POST)
    public ResponseEntity<String> updateEmployee(@RequestBody String employeeString) throws ParseException, Exception
    {
        try 
        {
            Employee emp;
            boolean isEmployeeManager = true;
            > //Logic to convert string to class obj using ObjectMapper;
            if(isEmployeeManager)
            {
                emp = mapper.readValue(employeeString, Manager.class);
            }
            else
            {
                emp = mapper.readValue(employeeString, Supervisor.class);
            }
            return employeeComponent.updateEmployee(emp);
        }
        catch(Exception e)
        {
            e.printStackTrace();
        }
    }
}

如何使用单个实体类和一个通用mongo存储库处理具有相同结构的两个不同的spring mongo集合

0 个答案: