Hibernate 5.2.17：ORA-01797：此运算符必须后跟ANY或ALL

Question

我有一个Oracle 18.4.0 XE数据库，我试图从JPA 2.1访问，由Hibernate 5.2.17实现。

我有两个实体之间的ManyToMany连接：

public class PermissionEntity implements Serializable {
    private static final long serialVersionUID = -3862680194592486778L;

    @Id
    @GeneratedValue
    private Long id;

    @Column(unique = true)
    private String permission;

    @ManyToMany
    private List<RoleEntity> roles;
}

public class RoleEntity implements Serializable {
    private static final long serialVersionUID = 8037069621015090165L;

    @Column(unique = true)
    private String name;

    @Id
    @GeneratedValue(strategy = GenerationType.SEQUENCE)
    private Long id;

    @ManyToMany(fetch = FetchType.LAZY, mappedBy = "roles")
    private List<PermissionEntity> permissions;
}

当试图在PermissionRepository：findAllByPermission(Iterable<String> permissions)上运行Spring Data JPA请求时，我得到以下异常：

Error : 1797, Position : 140, Sql = select permission0_.id as id1_0_, permission0_.permission as permission2_0_ from PermissionEntity permission0_ where permission0_.permission=(:1  , :2 ), OriginalSql = select permission0_.id as id1_0_, permission0_.permission as permission2_0_ from PermissionEntity permission0_ where permission0_.permission=(? , ?), Error Msg = ORA-01797: this operator must be followed by ANY or ALL

Answer 1

您告诉Spring Data Jpa引擎搜索Permission，其中权限等于列表。它应该使用IN运算符，因此您的方法名称应该是：

findAByPermissionIn(Iterable<String> permissions)

Answer 2

使用'in'关键字：findAllByPermissionIn(Iterable<String> permissions)。

这将生成如下查询：where permission0_.permission IN (:permissions)。