问题描述
我想在同一张桌子上的桌子上做一个简单的左联接。
public class User extends BaseEntity<User> {
public static final String PARENT_FIELD = "parent";
@OnetoOne(fetch = LAZY)
@JoinColumn(name = "parent_id")
User parent;
}
我在BaseEntity中有id列
我的标准代码是：
CriteriaBuilder cb = em.getCriteriaBuilder();
CriteriaQuery<Long> query = cb.createquery(Long.class);
Root<User> root = query.from(User.class);
Join<User,User> userJoin = root.join(PARENT_FIELD,JoinType.LEFT);
....some other joins
userJoin .on(
cb.equal(userJoin.getParent().get(PARENT_FIELD).get(ID_FIELD),root.get("id")),cb.equal(userJoin.get(ID_FIELD),root.get("id"))
);
....groupBy
...Heaving
问题是我在日志中看到此sql语句：
select
user0_.id as col_0_0_
from
user user0_
left outer join
user user2_
on user0_.parent_id=user2_.id !!!! this genrated from join!!! its excess. I just need
and ( only from sql!
user0_.parent_user_id=user0_.id <- this need or stament
and user2_.id=user0_.id
)
我想要的sql
select * from user p
left outer join user p1 on p.id = p1.parent_id or p.id = p1.id
我如何在没有条件的情况下制作标准鳕鱼？要么！！加入id字段，它也会帮助云... 我尝试在id上加入，但抛出了我无法在基本类型上加入的错误。 另一种方法是转换为select ... where或其他等效的条件...
解决方法
尝试一下：
ENTITY
public class User extends BaseEntity<User> {
public static final String PARENT_FIELD = "parent";
public static final String CHILDRENS_FIELD = "childrens";
@OneToOne(fetch = LAZY)
@JoinColumn(name = "parent_id")
User parent;
//Define bidirectional relation
@OneToMany(mappedBy = "parent",fetch = FetchType.LAZY,cascade = {})
private List<User> childrens;
}
标准
CriteriaBuilder cb = em.getCriteriaBuilder();
CriteriaQuery<Long> query = cb.createQuery(Long.class);
Root<User> root = query.from(User.class);
Join<User,User> userJoin = root.join(CHILDRENS_FIELD,JoinType.LEFT);
....some other joins
userJoin.on(
cb.or(cb.equal(userJoin.get(ID_FIELD),root.get("id")))
);
....groupBy
...Heaving
在定义双向关系时，初始化时创建的Join随您所希望的是User子级，而不是父级。
Join的on方法上，添加条件，以便将其封装在cb中。否则应根据需要生成查询。