Java JPQL 类似大小写不敏感
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JPQL Like Case Insensitive
提问by windupurnomo
I want to search data in User table by name case insensitive.
我想按名称不区分大小写搜索用户表中的数据。
@Repository
public interface UserRepository extends JpaRepository<User, Long> {
@Query("select u from User u where lower(u.name) like %lower(?1)%")
public List<User> findByNameFree(String name);
}
I got an error: unexpected token: %. Where should I place '%'?
我收到一个错误：意外令牌：%。我应该把“%”放在哪里？
采纳答案by Wim Deblauwe
You can use the concat operator:
您可以使用 concat 运算符：
@Query("select u from User u where lower(u.name) like lower(concat('%', ?1,'%'))")
public List<User> findByNameFree(String name);
or with a named parameter:
或使用命名参数：
@Query("select u from User u where lower(u.name) like lower(concat('%', :nameToFind,'%'))")
public List<User> findByNameFree(@Param("nameToFind") String name);
(Tested with Spring Boot 1.4.3)
（使用 Spring Boot 1.4.3 测试）
回答by Nicholas
You can use wildcard matching.
您可以使用wildcard matching.
for example, i want to search name like haha,
例如，我想搜索名称，例如haha，
@Query("select u from User u where lower(u.name) like :u_name")
public List<User> findByNameFree(@Param("u_name") String name);
List<User> users = userDao.findByNameFree("%haha");
回答by M. Deinum
If that is only what you want and you are using Spring Data JPA you don't need to write a query.
如果这只是您想要的并且您正在使用 Spring Data JPA，则不需要编写查询。
List<User> findByNameContainingIgnoreCase(String name);
Else you need to wrap the nameattribute with %before you pass it to the method (putting those directly in the query will simply not work). Or don't use a query but use a specification or the Criteria API to create the query.
否则，您需要在将name属性%传递给方法之前对其进行包装（将它们直接放在查询中是行不通的）。或者不使用查询，而是使用规范或 Criteria API 来创建查询。
