Java Spring数据JPA findAll，表在中间（PropertyReferenceException）

这是我的模型：

一个崇拜者与许多崇拜者联系在一起

一个Admis链接到0或一个admisret

实体：

public class AdmisHistory {
    @OneToMany(fetch = FetchType.EAGER, mappedBy = "admisHistory", cascade = CascadeType.ALL, orphanRemoval = true)
    private List<Admis> admis = new ArrayList<>();

public class Admis {
    @ManyToOne(cascade = CascadeType.ALL)
    @JoinColumn(name = ADMIS_HISTORY_ID)
    private AdmisHistory admisHistory;

    @OneToOne(mappedBy = "admis", cascade = CascadeType.ALL)
    private AdmisRejet admisRejet;

public class AdmisRejet {
    @OneToOne(cascade = CascadeType.ALL)
    @JoinColumn(name = ADMIS_ID)
    private Admis admis;

我尝试使用@Query，但语法不好：

public interface AdmisRejetRepository extends CrudRepository<AdmisRejet, Long> {
    @Query("SELECT a " +
            "FROM AdmisRejet ar " +
            "LEFT JOIN ar.admis a, " +
            "LEFT JOIN a.admisHistory ah " +
            "WHERE ah = :admisHistory")
    List< AdmisRejet> findAllByHistory(AdmisHistory admisHistory);
}

我怎么做呢？

试试看

  List< AdmisRejet> findAllAdmisRejetByAdmisHistoryWhereAdmisRejecIsNotNul(AdmisHistory admisHistory);

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或者类似的东西。

如果我正在阅读规范，请更正，这应该可以

@Query("SELECT DISTINCT ar 
        FROM AdmisRejet ar
        WHERE ar.admis.admisHistory = :history")
List<AdmisRejet> findAdmisRejetByAdmisHistory(AdmisHistory history);

---

使用@Query注释您的方法，并指定适当的JPQL查询作为注释的值。从方法名派生的自动查询只适用于像findByName这样的简单内容。@JBNizet不正确；Spring数据可以生成这样的遍历查询。问题在于关系模型，既因为admiss不是admis，也因为映射似乎不完全匹配。@chrylis admiss->admis打字错误。你的意思是“名单发现者管理者管理崇拜者；”生成横向查询的正确方法是什么？这个问题来自于我的实体？很难确切地说出对象图的含义，特别是因为不透明的联接表，但是你没有指定根对象，我认为你想要的是List findAllByAdmisAdmisRejetAdmisHistory。这就是说，由于反复使用Admis，直接用JPQL编写查询并将方法命名为可读的方法可能会更清晰。尝试使用@query，但找不到正确的语法请参见edit

public interface AdmisRejetRepository extends CrudRepository<AdmisRejet, Long> {
    @Query("SELECT a " +
            "FROM AdmisRejet ar " +
            "LEFT JOIN ar.admis a, " +
            "LEFT JOIN a.admisHistory ah " +
            "WHERE ah = :admisHistory")
    List< AdmisRejet> findAllByHistory(AdmisHistory admisHistory);
}

  List< AdmisRejet> findAllAdmisRejetByAdmisHistoryWhereAdmisRejecIsNotNul(AdmisHistory admisHistory);

@Query("SELECT DISTINCT ar 
        FROM AdmisRejet ar
        WHERE ar.admis.admisHistory = :history")
List<AdmisRejet> findAdmisRejetByAdmisHistory(AdmisHistory history);