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Spring Data JPA 关键字Exists的用法说明

作者：Jim~LoveQ

这篇文章主要介绍了Spring Data JPA 关键字Exists的用法说明，具有很好的参考价值，希望对大家有所帮助。如有错误或未考虑完全的地方，望不吝赐教

Spring Data JPA 关键字Exists

查询数据库中的此数据是否已存在：

例子：

查询sys_user表中的一个user是否存在，类SysUser对应的是数据库中的sys_user表，SysUserId是表sys_user的主键类（ID类）。

如果查询一个user，user的accountNo为demo。

userID为demo1，表sys_user的主键是accountNo和userID，下面代码中的方法是查询这个user是否存在，如果存在则返回true，不存在则返回false。

@Repository
public interface SysUserRepository extends JpaRepository<SysUser, SysUserId> {
    @Override
    boolean exists(SysUserId sysUserId);
}

Spring data jpa支持的关键字介绍

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

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)
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

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)

以上为个人经验，希望能给大家一个参考，也希望大家多多支持脚本之家。

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