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JPA方法名称中支持的关键字

JPA方法名称中支持的关键字

作者: 唯有努力不欺人丶 | 来源:发表于2019-08-26 14:59 被阅读0次

    额,这个是经常用到,但是每次查找很麻烦,记又没记住,所以从官网整理出来的,为了以后查找方便。

    关键词 样品 JPQL代码段
    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(附加参数绑定%)
    EndingWith findByFirstnameEndingWith … where x.firstname like ?1(与前置绑定的参数%)
    Containing findByFirstnameContaining … where x.firstname like ?1(包含参数绑定%)
    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> ages) … 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)

    出自spring官网spring-data-jpa官网

    原表截图
    原表截图

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