JPA error: The state field path '***' cannot be resolved to a valid type

Question

After running the following query method....   


@Query(value = "select s.owner, " +
            "count(case when s.status = 3 then 1 else null end) as agreed," +
            "count(case when s.status = 2 then 1 else null end) as rejected," +
            "count(case when s.status = 1 then 1 else null end) as pending " +
            "from Schedule s where (s.owner = ?1) " +
            "group by s.owner")

这里的状态是从枚举中获取的，我在运行时有一个转换器 class 用于 that.But 它给出了以下异常，即在转换器 class [=19= 上调用 convertToDatabaseColumn 时]$TypeConverter 的值为 1.e

有关此问题的任何帮助都将得到 helpful.Seek 专家的高度重视。

这是错误日志：

ReportQuery(referenceClass=Schedule jpql="select s.owner, count(case when s.status = 1 then 1 else null end) as agreed,count(case when s.status = 2 then 1 else null end) as rejected,count(case when s.status = 3 then 1 else null end) as pending from Schedule s where (s.owner = ?1) group by s.owner")
        at org.eclipse.persistence.exceptions.QueryException.prepareFailed(QueryException.java:1596)
        at org.eclipse.persistence.queries.DatabaseQuery.checkPrepare(DatabaseQuery.java:685)
        at org.eclipse.persistence.queries.ObjectLevelReadQuery.checkPrepare(ObjectLevelReadQuery.java:911)
        at org.eclipse.persistence.queries.DatabaseQuery.checkPrepare(DatabaseQuery.java:615)
        at org.eclipse.persistence.internal.jpa.EJBQueryImpl.buildEJBQLDatabaseQuery(EJBQueryImpl.java:194)
        at org.eclipse.persistence.internal.jpa.EJBQueryImpl.buildEJBQLDatabaseQuery(EJBQueryImpl.java:116)
        at org.eclipse.persistence.internal.jpa.EJBQueryImpl.<init>(EJBQueryImpl.java:102)
        at org.eclipse.persistence.internal.jpa.EJBQueryImpl.<init>(EJBQueryImpl.java:86)
        at org.eclipse.persistence.internal.jpa.EntityManagerImpl.createQuery(EntityManagerImpl.java:1614)
        at sun.reflect.GeneratedMethodAccessor86.invoke(Unknown Source)
        at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:43)
        at java.lang.reflect.Method.invoke(Method.java:498)
        at org.springframework.orm.jpa.ExtendedEntityManagerCreator$ExtendedEntityManagerInvocationHandler.invoke(ExtendedEntityManagerCreator.java:344)
        at com.sun.proxy.$Proxy75.createQuery(Unknown Source)
        at org.springframework.data.jpa.repository.query.SimpleJpaQuery.validateQuery(SimpleJpaQuery.java:86)
        ... 48 more
Caused by: javax.persistence.PersistenceException: An exception occurred while calling convertToDatabaseColumn on converter class com.orsbv.hcs.model.schedule.ScheduleStatus$TypeConverter with value 1

Answer 1

你应该做

更改s.status = 3（这应该是枚举）示例 s.status = Status.ACTIVE

或

变化s.status（应该是整数）= 3 示例：s.status.id = 3

Answer 2

通过使用完全限定的包名称可以解决这个问题。

@Query(value = "select s.owner, " + "count(case when s.status = com.orsbv.hcs.model.schedule.ScheduleStatus.AGREED then 1 else null end) as agreed," + "count(case when s.status = com.orsbv.hcs.model.schedule.ScheduleStatus.REJECTED then 1 else null end) as rejected," + "count(case when s.status = com.orsbv.hcs.model.schedule.ScheduleStatus.PENDING then 1 else null end) as pending " + "from Schedule s where (s.owner = ?1) " + "group by s.owner")

JPA error: The state field path '***' cannot be resolved to a valid type

JPA error: The state field path '***' cannot be resolved to a valid type

jpa

jpql

spring-data