您所做的一切都很好,尽管需要对 getUsersById 方法的返回类型进行小幅更正,它应该是 List<User> 而不是 List<Long>。
您可以使用 UserPKs 对象列表进行查询,并将其用作 IN 运算符的参数。所以下面的查询(就像你做的那样)可以正常工作:
public List<User> getUsersByIdCombo(List<UserPK> ids) {
TypedQuery<User> q = entityManager.createQuery("SELECT u FROM User AS u WHERE u.userPK IN :ids", User.class);
q.setParameter("ids", ids);
return q.getResultList();
}
您还可以通过显式遍历嵌入的 id 类(userPK.id 或 userPK.group)进行查询。以下是示例:
public List<User> getUsersById(List<Long> ids) {
TypedQuery<User> q = entityManager.createQuery("SELECT u FROM User AS u WHERE u.userPK.id IN :ids", User.class);
q.setParameter("ids", ids);
return q.getResultList();
}
public List<User> getUsersByGroup(List<String> groups) {
TypedQuery<User> q = entityManager.createQuery("SELECT u FROM User AS u WHERE u.userPK.group IN :groups", User.class);
q.setParameter("groups", groups);
return q.getResultList();
}
public User getSingleUserByIdAndGroup(Long id, String group) {
TypedQuery<User> q = entityManager.createQuery("SELECT u FROM User AS u WHERE u.userPK.id = :id AND u.userPK.group = :group", User.class);
q.setParameter("id", id);
q.setParameter("group", group);
return q.getSingleResult();
}
让我们通过测试每种方法来看看这些查询的行为。假设我们在您的 USER 表中有这些数据。
示例 1:通过 UserPK 对象列表查询用户
List<UserPK> userPKs = new ArrayList<UserPK>();
userPKs.add(new UserPK(1L, "it"));
userPKs.add(new UserPK(4L, "eng"));
List<User> userListQuery1 = userDAO.getUsersByIdCombo(userPKs);
for (User user: userListQuery1) {
System.out.println(user);
}
结果:
ID: 4 | GROUP: eng
ID: 1 | GROUP: it
示例 2:按 id 列表查询用户
List<Long> ids = new ArrayList<Long>();
ids.add(2L);
ids.add(5L);
List<User> userListQuery2 = userDAO.getUsersById(ids);
for (User user: userListQuery2) {
System.out.println(user);
}
结果:
ID: 5 | GROUP: eng
ID: 2 | GROUP: it
示例 3:按组列表查询用户
List<String> groups = new ArrayList<String>();
groups.add("it");
groups.add("eng");
List<User> userListQuery3 = userDAO.getUsersByGroup(groups);
for (User user: userListQuery3) {
System.out.println(user);
}
结果:
ID: 4 | GROUP: eng
ID: 5 | GROUP: eng
ID: 1 | GROUP: it
ID: 2 | GROUP: it
示例 4:按 id 和按组查询单个用户
User user = userDAO.getSingleUserByIdAndGroup(3L, "hr");
System.out.println(user);
结果:
ID: 3 | GROUP: hr