Question

我有一个要从JPA 2.1访问的Oracle 18.4.0 XE数据库，该数据库由Hibernate 5.2.17实现。

我在2个实体之间建立了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;
}

尝试在PermissionRepository：findAllByPermission(Iterable<String> permissions)上运行Spring Data JPA请求时，出现以下异常：

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

您正在告诉Spring Data Jpa引擎搜索权限等于列表的Permission。它应使用 IN 运算符，因此您的方法名称应为：

findAByPermissionIn(Iterable<String> permissions)

Answer 2

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

这将产生如下查询：where permission0_.permission IN (:permissions)。

休眠5.2.17：ORA-01797：此运算符后必须是ANY或ALL

2 个答案: