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[英]How to check if all Keys of the given Map are present in a one of the Maps contained within a List<Map<String, Object>> in Java efficiently?
[英]How to make a select statement on a list contained within an object in spring?
您如何在春季創建select語句或過濾嵌套在實體內的列表? 我有一個看起來像這樣的物體...
@Entity
@Table(name = "employee")
public class Employee {
...
@OneToMany(mappedBy = "_employee", fetch = FetchType.EAGER, cascade = CascadeType.REMOVE)
@JsonManagedReference
Set<Deal> _deals;
@OneToMany(mappedBy = "_employee", fetch = FetchType.EAGER, cascade = CascadeType.REMOVE)
@JsonManagedReference //This is simply to avoid a stackoverflow error according to this link http://stackoverflow.com/questions/3325387/infinite-recursion-with-jackson-json-and-hibernate-jpa-issue
Set<Recommendation> _recommendations;
@OneToMany(mappedBy = "_employee", fetch = FetchType.EAGER, cascade = CascadeType.REMOVE)
@JsonManagedReference //This is simply to avoid a stackoverflow error according to this link http://stackoverflow.com/questions/3325387/infinite-recursion-with-jackson-json-and-hibernate-jpa-issue
Set<Event> _events;
public Employee() {
}
//getters and setters
....
我為員工提供了一個由服務類訪問的存儲庫。
存儲庫如下所示。
public interface EmployeeRepository extends CrudRepository<Employee, Long> {
public Employee getEmployeeById(Long _id);
public Employee getEmployeeBy_username(String username);
}
因此,特別是當我通過其ID獲取一名雇員時,它會返回上述列表。 找回員工時,我需要以某種方式進行選擇語句或過濾器_deals,_recommendations和_events。 這樣只有那些具有boolean屬性_active = true的人才返回。 到目前為止,無論是否有效,都會返回所有交易建議和事件。 如何過濾或從這些列表中僅選擇活動對象?
您幾乎總是為每個查詢選擇一個Entity類型,最好在數據庫中進行過濾。 如果您希望交易,推薦和事件屬於特定的雇員,通常我會將這些方法放在屬於我要加載的實體類型的存儲庫中,如下所示:
@Repository
public interface DealRepository extends JpaRepository<Deal, Long> {
@Query("select d from Deal d where d.active= true and d.employee.id = :employeeId")
List<Deal> findActiveDeals(@Param("employeeId") long employeeId);
}
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