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JPA 常用方法

JPA 常用方法

作者: 凯睿看世界 | 来源:发表于2019-01-06 15:13 被阅读29次

    具体的关键字,使用方法和生产成SQL如下表所示

    Keyword Sample JPQL snippet
    And findByLastnameAndFirstname … where x.lastname = ?1 and x.firstname = ?2
    Or findByLastnameOrFirstname … where x.lastname = ?1 or x.firstname = ?2
    Is,Equals 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 ages) … where x.age in ?1
    NotIn findByAgeNotIn(Collection 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)

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