有没有办法在以下查询中使用 HQL？
SELECT userId, pwd, pwdDate FROM
(SELECT userId, AES_DECRYPT(pwd, 'key_str') as pwd, pwdDate
FROM UserHistory order by pwdDate desc limit 5 ) AS A
WHERE pwd = :pwd
以下有效。
SELECT *
FROM UserHistory order by pwdDate desc limit 5
上面的sql在hibernate中可以是下面的
Criteria criteria = session.createCriteria(UserHistory.class);
criteria.addOrder(Order.desc("pwdDate"));
List<UserHistory> list = criteria.setMaxResults(5).list();
最佳答案
以下工作。关键是为 Native SQL 创建 SQLQuery。
String SQL =
"SELECT A.* FROM \n" +
"(select * \n" +
" from user_history$ order by pwdDate desc limit 5 ) AS A \n" +
"where pwd = AES_ENCRYPT(:pwd, 'key_str') \n";
Query query = session.createSQLQuery(SQL);
query.setParameter("pwd", psw);
List<UserHistory> list = query.list();
关于java - Hibernate 或 HQL 选择另一个 SELECT 的结果，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/47426317/