首页 > 代码库 > spring jpa 语法

spring jpa 语法

2024-09-08 15:30:44   219人阅读

摘自http://www.cnblogs.com/BenWong/p/3890012.html

Table 2.3. Supported keywords inside method names

KeywordSampleJPQL snippet
And findByLastnameAndFirstname … where x.lastname = ?1 and x.firstname = ?2
Or findByLastnameOrFirstname … where x.lastname = ?1 or x.firstname = ?2
Is,Equals findByFirstname,findByFirstnameIs,findByFirstnameEquals … where x.firstname = 1?
Between findByStartDateBetween … where x.startDate between 1? and ?2
LessThan findByAgeLessThan … where x.age < ?1
LessThanEqual findByAgeLessThanEqual … where x.age <= ?1
GreaterThan findByAgeGreaterThan … where x.age > ?1
GreaterThanEqual findByAgeGreaterThanEqual … where x.age >= ?1
After findByStartDateAfter … where x.startDate > ?1
Before findByStartDateBefore … where x.startDate < ?1
IsNull findByAgeIsNull … where x.age is null
IsNotNull,NotNull findByAge(Is)NotNull … where x.age not null
Like findByFirstnameLike … where x.firstname like ?1
NotLike findByFirstnameNotLike … where x.firstname not like ?1
StartingWith findByFirstnameStartingWith … where x.firstname like ?1 (parameter bound with appended %)
EndingWith findByFirstnameEndingWith … where x.firstname like ?1 (parameter bound with prepended %)
Containing findByFirstnameContaining … where x.firstname like ?1 (parameter bound wrapped in %)
OrderBy findByAgeOrderByLastnameDesc … where x.age = ?1 order by x.lastname desc
Not findByLastnameNot … where x.lastname <> ?1
In findByAgeIn(Collection<Age> ages) … where x.age in ?1
NotIn findByAgeNotIn(Collection<Age> age) … where x.age not in ?1
True findByActiveTrue() … where x.active = true
False findByActiveFalse() … where x.active = false
IgnoreCase findByFirstnameIgnoreCase … where UPPER(x.firstame) = UPPER(?1)
 
 

spring jpa 语法


联系
我们