hibernate3学习笔记(七) Criteria Queries - 编程入门网
作者 佚名技术
来源 NET编程
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发布时间 2012-06-20
_.age as age0_0_ from user this_ order by this_.age ascid name/age11 yangye/null12 yangye/null9 yangye/2310 yangye/232 chenyan/261 shenbin/28Hibernate: select this_.id as id0_0_, this_.name as name0_0_, this_.age as age0_0_ from user this_ where this_.age>? and this_.age<?id name/age2 chenyan/269 yangye/2310 yangye/23Hibernate: select this_.id as id0_0_, this_.name as name0_0_, this_.age as age0_0_ from user this_ where (this_.age=? or this_.age is null)id name/age9 yangye/2310 yangye/2311 yangye/null12 yangye/nullHibernate: select this_.id as id0_0_, this_.name as name0_0_, this_.age as age0_0_ from user this_ where this_.name LIKE (?)id name/age2 chenyan/269 yangye/2310 yangye/2311 yangye/null12 yangye/nullHibernate: select this_.id as id0_0_, this_.name as name0_0_, this_.age as age0_0_ from user this_ where this_.age BETWEEN (?) AND (?)id name/age1 shenbin/282 chenyan/26Hibernate: select this_.id as id0_0_, this_.name as name0_0_, this_.age as age0_0_ from user this_ limit ?, ?id name/age10 yangye/2311 yangye/nullHibernate: select count(*) as y0_ from user this_6Hibernate: select avg(this_.age) as y0_ from user this_25.0Hibernate: select this_.age as y0_ from user this_ group by this_.agenull232628Hibernate: select this_.age as y0_, this_.name as y1_, count(*) as y2_ from user this_ group by this_.age, this_.namenull yangye 223 yangye 226 chenyan 128 shenbin 1Hibernate: select this_.id as id0_0_, this_.name as name0_0_, this_.age as age0_0_ from user this_ where this_.name like ? order by this_.age descid name/age2 chenyan/269 yangye/2310 yangye/2311 yangye/null12 yangye/nullHibernate: select this_.id as id0_0_, this_.name as name0_0_, this_.age as age0_0_ from user this_ where (this_.age=?)id name/age2 chenyan/26
hibernate3学习笔记(七) Criteria Queries(3)时间:2011-02-02 csdn博客 沈斌注意： Restrictions的几个常用限定查询方法如下表所示：
此外，Criteria还可以进行复合查询。即在原有的查询基础上再进行查询，例如 |
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