Question

我在hibernate中有@manytomany关系喜欢：表员工

public class Employee implements Serializable {
    @ManyToMany(fetch = FetchType.EAGER)
    @Fetch(FetchMode.SELECT)
    @JoinTable(name = "employee_role", joinColumns = { @JoinColumn(name = "employee_id") }, inverseJoinColumns = { @JoinColumn(name = "role_id") })
    private Set<Role> roles = new HashSet<Role>(0);
}

表格角色：

public class Role implements Serializable {
    @Id
    @GeneratedValue(strategy = GenerationType.SEQUENCE)
    @Column(name = "role_id", unique = true, nullable = false)
    @Basic(fetch = FetchType.EAGER)
    private long id;
}

如何通过 hibernate Criterion 在角色具有特定值时获得所有员工或子查询

Answer 1

如果我理解正确，您希望让所有员工都具有特定的角色。为什么要使用Criteria API。 HQL更简单易读：

select e from Employee e inner join e.roles role where role.id = :roleId

如果您真的想使用Criteria API，请点击此处：

Criteria c = session.createCriteria(Employee.class, "employee");
c.createAlias("employee.roles", "role");
c.add(Restrictions.eq("role.id", roleId));
List<Employee> employeed = c.list();

Answer 2

尝试JB Nizet的回答并在表中返回多行不同的数字解决方案是添加setResultTransformer（Criteria.DISTINCT_ROOT_ENTITY）

Criteria c = getCurrentSession().createCriteria(Employee.class,"employee");
c.createAlias("employee.roles", "role");
c.add(Restrictions.eq("role.id", id));
List<Employee> employeed = c.setResultTransformer(Criteria.DISTINCT_ROOT_ENTITY).list();

manytomany检索标准

2 个答案: