CriteriaQuery 中的子查询答案

【问题标题】：SubQuery in a CriteriaQueryCriteriaQuery 中的子查询
【发布时间】：2020-09-10 06:26:46
【问题描述】：

我有抽象类 Entity 和这两个实体 Role 和 User 从实体扩展：

@Entity@Inheritance(strategy = InheritanceType.TABLE_PER_CLASS)
public abstract class Entity {

    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    protected long id = 0; }
    

@Entity
public class Role extends Entity{
    @Column(name = "NAME", unique = true, nullable = false)
    private String name;
}

@Entity
public class User extends Entity{
    @Column(name = "ABBREVIATION", unique = true, nullable = false)
    private String abbreviation;

    //The user can have several roles
    @ManyToMany
    @JoinTable(
        name = "USER_ROLE",
        joinColumns = @JoinColumn(name = "USER_ID"),
        inverseJoinColumns = @JoinColumn(name = "ROLE_ID"))
        private Set<Role> roles = new HashSet<>();
}

现在，当我单击角色时，我希望显示用户的缩写。它的本机查询是：

query = em.createNativeQuery("SELECT ABBREVIATION FROM dbtest.user WHERE ID IN"
                        + " (SELECT USER_ID FROM dbtest.user_role WHERE ROLE_ID = " + role.getId() + ")");

我现在已经创建了一个条件查询，它总是给我一个空列表。

CriteriaBuilder cb = em.getCriteriaBuilder();
CriteriaQuery<Role> cq = cb.createQuery(Role.class);
Root<User> root = cq.from(User.class);

Subquery subquery = cq.subquery(Role.class);
Root subfrom = subquery.from(Role.class);

subquery.select(subfrom.get(Role_.id));
subquery.where(cb.equal(subfrom.get(Role_.id), role.getId()));

cq.multiselect(root.get(User_.abbreviation));
cq.where(cb.equal(root.get(User_.id), subquery));

query = em.createQuery(cq);
return query.getResultList();

这里有什么问题吗？有人可以帮忙吗？

【问题讨论】：

你检查日志中创建的actual query了吗？
@Sujitmohanty30 天哪，我明白了：从用户中选择新实体。 )
查看您的子查询。它会生成像"SELECT ROLE_ID FROM user_role WHERE ROLE_ID = " + role.getId() 这样的无用查询。所以它返回role.getId() 值。您必须在子查询中选择 USER_ID，但您的实体不包含此类字段。我已经在我的回答中说明了

标签： java sql jpa criteria criteria-api

【解决方案1】：

首先，设置正确的类型：

CriteriaBuilder cb = em.getCriteriaBuilder();
//CriteriaQuery<Role> cq = cb.createQuery(Role.class);
//The type of the data returned in multiselect.
CriteriaQuery<String> cq = cb.createQuery(String.class);
Root<User> root = cq.from(User.class);
//Add to compare
Join<User,Role> joinRole = root.join(User_.roles,JoinType.Inner);

//The type of the data returned in select.
//Subquery subquery = cq.subquery(Role.class);
Subquery<Long> subquery = cq.subquery(Long.class);
//Root subfrom = subquery.from(Role.class);
//We set the data type in the Root
Root<Role> subfrom = subquery.from(Role.class);

subquery.select(subfrom.get(Role_.id));
subquery.where(cb.equal(subfrom.get(Role_.id), role.getId()));

cq.multiselect(root.get(User_.abbreviation));
//cq.where(cb.equal(root.get(User_.id), subquery));
//To use in the where subquery, call the getSelection () method
cq.where(cb.equal(root.get(User_.id), subquery.getSelection()));

query = em.createQuery(cq);
return query.getResultList();

另外，主查询的 where 是 in 不是 equals，改变这个

cq.where(cb.equal(root.get(User_.id), subquery.getSelection()));

通过这个

cq.where(joinRole.get(Role_.id).in(subquery.getSelection()));

【讨论】：

谢谢，我得到了这个查询：select new java.lang.String(generatedAlias0.abbreviation) from User as generatedAlias0 where generatedAlias0.id in ((select generatedAlias1.id from Role as generatedAlias1 where generatedAlias1.id =12L)) hibernate 生成的查询不一样，我需要 user_role 表中的 user_id，在这个查询中，我在第二个中选择 role_id from role
我编辑我的代码，在主查询中创建 Join 并将 cq.where 更改为 cq.where(joinRole.get(Role_.id).in(subquery.getSelection()));

【解决方案2】：

CriteriaBuilder cb = em.getCriteriaBuilder();
CriteriaQuery<Role> cq = cb.createQuery(Role.class);
Root<User> root = cq.from(User.class);

Subquery<Long> subquery = cq.subquery(Long.class);
Root<Role> subqueryRoot = subquery.from(Role.class);
Predicate subqueryPredicate = 
        cb.equal(subqueryRoot.get(Role_.id), role.getId());

//!!! I suspect here you have to select USER_ID, but Role entity does not contain such field        
subquery.select(subqueryRoot.get(Role_.id)) 
        .where(subqueryPredicate); 

Predicate queryPredicate = cb.in(root.get(User_.id)).value(subquery); 
cq.multiselect(root.get(User_.abbreviation)).where(queryPredicate);

【讨论】：