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JPA Repository方法名查询支持语法

JPA Repository方法名查询支持语法

作者: 湘西刺客王胡子 | 来源:发表于2021-04-27 16:19 被阅读0次

    搬运自官方文档: Spring Data JPA - Reference Documentation

    Keyword Sample JPQL snippet
    Distinct findDistinctByLastnameAndFirstname select distinct … where x.lastname = ?1 and x.firstname = ?2
    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, Null findByAge(Is)Null … 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> ages) … where x.age not in ?1
    TRUE findByActiveTrue() … where x.active = true
    FALSE findByActiveFalse() … where x.active = false
    IgnoreCase findByFirstnameIgnoreCase … where UPPER(x.firstname) = UPPER(?1)

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