hibernate根据字段去除重复记录
public List<Employee> getUniqueNameByHqlCorrect(){
??List<Object[]> list=employeeDao.createQuery("from Employee e inner join (select max(b.id) as id? from? Employee b group by b.name) c on (c.id=e.id)", new Object[]{}).list();
??for(Object[] rows:list){??
??????????? System.out.println(": " + rows[0] + "值: " + rows[1]);????
??????? }?
??return null;
?}
from Employee e inner join (select max(b.id) as id? from? Employee b group by b.name) c on (c.id=e.id)
这条hql会不会写错,为什么会报.异常
org.hibernate.hql.ast.QuerySyntaxException: unexpected token: ( near line 1, column 57 [from com.dhhc.cs.oa.entity.system.Employee e inner join (select max(b.id) as id? from? com.dhhc.cs.oa.entity.system.Employee b group by b.name) c on c.id=e.id]
?at
?
?
?
以下两种方法可以实现根据字段去除重复记录
?
?
/**
? * hibernate sql根据某一字段的值不重复,用in的方法不健壮，效率低，当in里的id值大于1000个报错
? * @return
? */
?public List<Employee> getUniqueNameByHqlCommon(){
??List<Employee> list=employeeDao.createQuery("from Employee e where e.id in(select max(b.id) from? Employee b group by b.name)", new Object[]{}).list();
??for(Employee entity:list){??
???System.out.println(": " +entity.getName() );????
??}?
??return null;
?}
?
?
?
/**
? * 使用联接,相对于in的用法更合适
? * @return
? */
?public List<Employee> getUniqueNameBySqlCorrect(){
??Session session=this.getSessionFactory().openSession();
??String sql="select name,email from sys_employee e inner join (select max(b.id) as id from? sys_employee b group by b.name) c on (c.id=e.id)";
??List<Object[]> list=session.createSQLQuery(sql).list();
??for(Object[] rows:list){??
??????????? System.out.println(": " + rows[0] + "值: " + rows[1]);????
??????? }
??session.close();
??return null;
?}
?
?
