引起：org.hibernate.QueryException：无法取消引用标量集合元素：角色

Question

我有这门课，我正在尝试从用户表中访问 mobileNumbers 和 roles，但是被 lazy 初始化后，我不断收到 'lazy intialization error。

我不想删除LAZY fetch，因为很少需要完全访问该对象，但这是必需的。所以为了克服它，我首先尝试添加@Transactional，但通过这篇文章https://codete.com/blog/5-common-spring-transactional-pitfalls/，在我的情况下这似乎是一个糟糕的方法，我尝试使用join fetch，但它一直给multiplebagfetchexception，因此我试图获取它们一次（我的意思是元素集合）

使用这个存储库类

public interface UserRespository extends JpaRepository<UserDao, Long> {

    Optional<UserDao> getByUserNameIgnoreCase(String userName);
//    Optional<UserDao> findByUserNameIgnoreCase(String userName);

    @Query(value = "select dao.roles.roles from UserDao dao inner join dao.roles r on dao.userName in elements(r.userName) and upper(dao.userName) = upper(?1)")
    Object getByUserNameIgnoreCaseComplete2(String userName);
}

然后我得到他的错误

Caused by: org.hibernate.QueryException: cannot dereference scalar collection element: roles
    at org.hibernate.persister.collection.ElementPropertyMapping.toType(ElementPropertyMapping.java:33) ~[hibernate-core-5.3.9.Final.jar:5.3.9.Final]
    at org.hibernate.persister.collection.AbstractCollectionPersister.toType(AbstractCollectionPersister.java:1644) ~[hibernate-core-5.3.9.Final.jar:5.3.9.Final]
    at org.hibernate.hql.internal.ast.tree.FromElementType.getPropertyType(FromElementType.java:396) ~[hibernate-core-5.3.9.Final.jar:5.3.9.Final]
    at org.hibernate.hql.internal.ast.tree.FromElement.getPropertyType(FromElement.java:515) ~[hibernate-core-5.3.9.Final.jar:5.3.9.Final]
    at org.hibernate.hql.internal.ast.tree.DotNode.getDataType(DotNode.java:682) ~[hibernate-core-5.3.9.Final.jar:5.3.9.Final]
    at org.hibernate.hql.internal.ast.tree.DotNode.prepareLhs(DotNode.java:265) ~[hibernate-core-5.3.9.Final.jar:5.3.9.Final]
    at org.hibernate.hql.internal.ast.tree.DotNode.resolve(DotNode.java:205) ~[hibernate-core-5.3.9.Final.jar:5.3.9.Final]
    at org.hibernate.hql.internal.ast.tree.FromReferenceNode.resolve(FromReferenceNode.java:114) ~[hibernate-core-5.3.9.Final.jar:5.3.9.Final]
    at org.hibernate.hql.internal.ast.tree.FromReferenceNode.resolve(FromReferenceNode.java:109) ~[hibernate-core-5.3.9.Final.jar:5.3.9.Final]
    at org.hibernate.hql.internal.ast.tree.FromReferenceNode.resolve(FromReferenceNode.java:104) ~[hibernate-core-5.3.9.Final.jar:5.3.9.Final]
    at org.hibernate.hql.internal.ast.tree.DotNode.resolveSelectExpression(DotNode.java:744) ~[hibernate-core-5.3.9.Final.jar:5.3.9.Final]
    at org.hibernate.hql.internal.ast.HqlSqlWalker.resolveSelectExpression(HqlSqlWalker.java:1057) ~[hibernate-core-5.3.9.Final.jar:5.3.9.Final]
    at org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.selectExpr(HqlSqlBaseWalker.java:2295) ~[hibernate-core-5.3.9.Final.jar:5.3.9.Final]
    at org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.selectExprList(HqlSqlBaseWalker.java:2232) ~[hibernate-core-5.3.9.Final.jar:5.3.9.Final]
    at org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.selectClause(HqlSqlBaseWalker.java:1503) ~[hibernate-core-5.3.9.Final.jar:5.3.9.Final]
    at org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.query(HqlSqlBaseWalker.java:585) ~[hibernate-core-5.3.9.Final.jar:5.3.9.Final]
    at org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.selectStatement(HqlSqlBaseWalker.java:313) ~[hibernate-core-5.3.9.Final.jar:5.3.9.Final]
    at org.hibernate.hql.internal.antlr.HqlSqlBaseWalker.statement(HqlSqlBaseWalker.java:261) ~[hibernate-core-5.3.9.Final.jar:5.3.9.Final]
    at org.hibernate.hql.internal.ast.QueryTranslatorImpl.analyze(QueryTranslatorImpl.java:271) ~[hibernate-core-5.3.9.Final.jar:5.3.9.Final]
    at org.hibernate.hql.internal.ast.QueryTranslatorImpl.doCompile(QueryTranslatorImpl.java:191) ~[hibernate-core-5.3.9.Final.jar:5.3.9.Final]
    ... 112 common frames omitted

我不知道为什么我一直收到这个错误。有没有其他方法可以获得整个对象？？

请帮忙。

实际类

@Entity
@Table(name = "users")
@Data
@NoArgsConstructor
public class UserDao implements Serializable {

    @Id
    @GeneratedValue
    private long id;

    @Column(name = "user_name", nullable = false, unique = true)
    private String userName;

    @Column(name = "password", nullable = false)
    private String password;

    @ElementCollection(targetClass = java.lang.String.class)
    @CollectionTable(name = "mobile_numbers_List",joinColumns = @JoinColumn(name = "user_name",referencedColumnName = "user_name"))
    @Column(name = "mobile_number")
    private List<String> mobileNumbers;

    @ElementCollection(targetClass = java.lang.String.class)
    @CollectionTable(name = "roles_list",joinColumns = @JoinColumn(name = "user_name",referencedColumnName = "user_name"))
    @Column(name = "roles")
    private List<String> roles;
}

Answer 1

这个问题好像你previous questio的问题，上一个和这个的答案是一样的。

您的查询在这里是错误的

    @Query(value = "select dao.roles.roles from UserDao dao inner join dao.roles r on dao.userName in elements(r.userName) and upper(dao.userName) = upper(?1)")
    Object getByUserNameIgnoreCaseComplete2(String userName);

它正在尝试获取 dao.roles.roles，如果它甚至不存在。

这里有一些你应该尝试的改变

1. 把dao.roles.roles改成r（因为你已经创建了别名，让我们使用它）
2. 删除dao.userName in elements(r.userName) 不是必需的，spring 会自动执行此操作，甚至无需指定它。
3. 将返回类型更改为 list ，它是 ElementCollection 即 Collection 元素

这就是您的最终查询的样子

    @Query(value = "select r from UserDao dao inner join dao.roles r on upper(dao.userName) = upper(?1)")
    List<String> getByUserNameIgnoreCaseComplete2(String userName);

Answer 2

你为什么要评论 findByUserNameIgnoreCase ？ findBy为我工作 Take a peek in this page

Answer 3

问题是您查询中的dao.roles.roles。将其更改为dao.roles。 dao.roles 在您的情况下是 List<String> 而 String 没有属性 roles。