Question

我正在开发一个Spring-MVC应用程序，我有一个多对多关系，我必须在2个表中查询以获取我需要的值。我会更详细地解释一下。

我有2个表GroupAccount，GroupMembers有多对多关系。现在有一个名为membertable的联结表存储来自GroupMembers和GroupAccount的ID。

这就是我要找的：

我传递了groupAccounId和username作为参数。现在，在 GroupMembers表，存储了一个用户名。在groupAccount中，存储了groupAccountId。
现在在memberjunction中，我有复合键 memberid，GroupAccountId，我想要用户名的成员ID 我提交了一个匹配的groupAccountId。

下面是SQL代码和Spring-mvc代码，以便更好地理解。

CREATE TABLE public.groupaccount (
                groupid NUMERIC NOT NULL,
                groupname VARCHAR,

                groupaccountstatus BOOLEAN DEFAULT false NOT NULL,
                adminusername VARCHAR,
                CONSTRAINT groupid PRIMARY KEY (groupid)
);

CREATE TABLE public.groupmembers (
                memberid INTEGER NOT NULL,
                musername VARCHAR
                CONSTRAINT memberid PRIMARY KEY (memberid)
);
CREATE TABLE public.memberjunction (
                memberid INTEGER NOT NULL,
                groupid NUMERIC NOT NULL,
                CONSTRAINT membergroupid PRIMARY KEY (memberid, groupid)
);

GroupMembersDAOImpl：＃

 @Override
    public List<Integer> returnMemberIdWithMatchingUsername(String memberUsername) {
        session = this.sessionFactory.getCurrentSession();
        org.hibernate.Query query = session.createQuery("From GroupMembers as " +
                "n where n.memberUsername=:memberUsername");
        query.setParameter("memberUsername",memberUsername);
        List<GroupMembers> memberList = query.list();
        List<Integer> memberIdList = new ArrayList<>();
        for(GroupMembers members :memberList){
                memberIdList.add(members.getMemberid());
        }
        return memberIdList;
    }

GroupAccount模型：

@Entity
@Table(name="groupaccount")
public class GroupAccount {

@Id
    @Column(name="groupid")
    @GeneratedValue(strategy = GenerationType.SEQUENCE,generator = "groupaccount_seq_gen")
    @SequenceGenerator(name = "groupaccount_seq_gen",sequenceName = "groupaccount_seq")
    private Long groupId;

 @ManyToMany(cascade = CascadeType.ALL)
    @JoinTable(name = "memberjunction", joinColumns = {@JoinColumn(name = "groupid")},
                inverseJoinColumns = {@JoinColumn(name = "memberid")})
    private Set<GroupMembers> groupMembersSet = new HashSet<>();

    public void setGroupMembersSet(Set<GroupMembers> groupMembersSet){
        this.groupMembersSet = groupMembersSet;
    }
}

GroupMembers模型类：

@Entity
@Table(name="groupmembers")
public class GroupMembers {
 @Id
    @Column(name="memberid")
    @GeneratedValue(strategy = GenerationType.SEQUENCE,generator = "groupmembers_seq_gen")
    @SequenceGenerator(name = "groupmembers_seq_gen",sequenceName = "groupmembers_seq")
    private int memberid;

    @ManyToMany(mappedBy = "groupMembersSet")
    private Set<GroupAccount> groupAccounts = new HashSet<>();

    public void setGroupAccounts(Set<GroupAccount> groupAccounts){
        this.groupAccounts = groupAccounts;
    }

    public Set<GroupAccount> getGroupAccounts(){
        return this.groupAccounts;
    }
}

我正在使用的查询：

 @Override
    public int getMemberIdForCanvas(String memberUsername, Long groupId) {
        session = this.sessionFactory.getCurrentSession();
        org.hibernate.Query query = session.createQuery("select distinct m.memberId from GroupMembers m\n" +
                "join m.groupAccounts a\n" +
                "where a.memberUsername = :userName and m.groupId=:groupId");

        query.setParameter(memberUsername,"memberUsername");
        query.setParameter(String.valueOf(groupId),"groupId");
        int memberid = (Integer)query.uniqueResult();
        return memberid;
    }

任何帮助都会很好。非常感谢。

Answer 1

此处the documentation for joins and HQL。请阅读。

查询就像

一样简单

select distinct m.memberId from GroupMembers m
join m.groupAccounts a
where a.memberUsername = :userName

还请修复您的命名。 GroupMembers实例是单个组成员。因此，该类应命名为GroupMember，而不是s。在此类的字段中重复该类的名称也是多余的：member.getId()比member.getMemberId()更易读，更简洁。其他领域也一样。

在Hibernate中查询复合表

1 个答案: