JPA/JPA

조건문 매칭

lovineff 2020. 12. 1. 12:35

아래 표는 JPA 사용시 WHERE 조건절을 매칭을 나타낸다.

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)

 

출처 : 공식 문서

docs.spring.io/spring-data/jpa/docs/current/reference/html/#reference

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