hibernate_Restrictions用法

hibernate_Restrictions用法
Restrictions.eq –> equal,等于.
Restrictions.allEq –> 参数为Map对象,使用key/value进行多个等于的比对,
相当于多个Restrictions.eq的效果
Restrictions.gt –> great-than > 大于
Restrictions.ge –> great-equal >= 大于等于
Restrictions.lt –> less-than, < 小于
Restrictions.le –> less-equal <= 小于等于
Restrictions.between –> 对应SQL的between子句
Restrictions.like –> 对应SQL的LIKE子句
Restrictions.in –> 对应SQL的in子句
Restrictions.and –> and 关系
Restrictions.or –> or 关系
Restrictions.isNull –> 判断属性是否为空,为空则返回true
Restrictions.isNotNull –> 与isNull相反
Restrictions.sqlRestriction –> SQL限定的查询
Order.asc –> 根据传入的字段进行升序排序
Order.desc –> 根据传入的字段进行降序排序
MatchMode.EXACT –> 字符串精确匹配.相当于”like ‘value’”
MatchMode.ANYWHERE –> 字符串在中间匹配.相当于”like ‘%value%’”
MatchMode.START –> 字符串在最前面的位置.相当于”like ‘value%’”
MatchMode.END –> 字符串在最后面的位置.相当于”like ‘%value’”

例子
查询年龄在20-30岁之间的所有学生对象
List list = session.createCriteria(Student.class)
.add(Restrictions.between("age",new Integer(20),new Integer(30)).list();
查询学生姓名在AAA,BBB,CCC之间的学生对象
String[] names = {"AAA","BBB","CCC"};
List list = session.createCriteria(Student.class)
.add(Restrictions.in("name",names)).list();
查询年龄为空的学生对象
List list = session.createCriteria(Student.class)
.add(Restrictions.isNull("age")).list();
查询年龄等于20或者年龄为空的学生对象
List list = session.createCriteria(Student.class)
.add(Restrictions.or(Restrictions.eq("age",new Integer(20)),
Restrictions.isNull("age")).list();

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使用QBC实现动态查询 
public List findStudents(String name,int age){

Criteria criteria = session.createCriteria(Student.class);
if(name != null){
criteria.add(Restrictions.liek("name",name,MatchMode.ANYWHERE));
}
if(age != 0){
criteria.add(Restrictions.eq("age",new Integer(age)));
}
criteria.addOrder(Order.asc("name"));//根据名字升序排列
return criteria.list();
}