Hibernate hql: 只支持用法 INSERT INTO ... SELECT ... FROM

下面是两个例子：

 Query query=session.createQuery("insert into UserInfo(username,password,email) values(?,?,?)");

像上面的hql语句在编译的时候会报错，

 Exception in thread "main" org.hibernate.hql.ast.QuerySyntaxException: expecting OPEN, found 'values' near line 1, column 22 [insert into Userinfo values('aaf')]
     at org.hibernate.hql.ast.QuerySyntaxException.convert(QuerySyntaxException.java:54)
     at org.hibernate.hql.ast.QuerySyntaxException.convert(QuerySyntaxException.java:47)
     at org.hibernate.hql.ast.ErrorCounter.throwQueryException(ErrorCounter.java:82)
     at org.hibernate.hql.ast.QueryTranslatorImpl.parse(QueryTranslatorImpl.java:284)
     at org.hibernate.hql.ast.QueryTranslatorImpl.doCompile(QueryTranslatorImpl.java:182)
     at org.hibernate.hql.ast.QueryTranslatorImpl.compile(QueryTranslatorImpl.java:136)
     at org.hibernate.engine.query.HQLQueryPlan.(HQLQueryPlan.java:101)
     at org.hibernate.engine.query.HQLQueryPlan.(HQLQueryPlan.java:80)
     at org.hibernate.engine.query.QueryPlanCache.getHQLQueryPlan(QueryPlanCache.java:94)
     at org.hibernate.impl.AbstractSessionImpl.getHQLQueryPlan(AbstractSessionImpl.java:156)
     at org.hibernate.impl.AbstractSessionImpl.createQuery(AbstractSessionImpl.java:135)
     at org.hibernate.impl.SessionImpl.createQuery(SessionImpl.java:1651)
     at com.xmetc.hibernate.test.HQL07.main(HQL07.java:23)

因此，

 //String queryString3 = "insert into Userinfo  values()";//不支持.

 String queryString3 = "insert into Userinfo select ... from .. ";//支持.