@Query("select e from Meeting e where e.meetingName like %:query%" )
2.@Query( nativeQuery = true, value ="select * from meeting as m where m.meeting_name like &
我想知道是否有可能在不使用@Query注释的情况下创建一个JPA,以便从我正在查询的对象内部的对象属性中获取结果。Parameter value [1] did not match expected type [project.model.oc.Ocorrencia (n/a)] 我已经使用这里的https://docs.spring.io/spring-data/jpa/docs/current/reference/
at org.springframework.data.jpa.repository.query.PartTreeJpaQuery.<init>(PartTreeJpaQuery.java:96) ~[spring-data-jpa-2.5.4.jar:2.5.4] at org.springframewo
at org.springframework.data.jpa.repository.query.ParameterBinder.bind(ParameterBinder.java:111) ~[spring-data-jpa) ~[spring-data-jpa-1.7.2.RELEASE.jar:?]~[spring-data-jpa-1.7.2.RELEASE.jar:?]~[spring-data-jpa-1