如何使用JPA Criteria API在左連接上指定多個條件？

Question

我想轉換以下SQL查詢：

select * from region_tree country left outer join region_tree region   
on country.REG_CODE_PAR=region.REG_CODE  
and region.LFT < country.LFT   
and region.RGT > country.RGT  
and region.REG_CODE_PAR = 'ALL'  
and COUNTRY.STATUS_CODE = 'A'  
and REGION.STATUS_CODE = 'A  

進入基於JPA Crtieria的查詢。

我創建了一個表示自聯接的實體：

@Entity  
@Table(name = "REGION_TREE")  
public class RegionTree implements Serializable {  
    ... some other attributes  

    @ManyToOne  
    @JoinColumn(name = "REG_CODE_PAR")  
    private RegionTree region;   

    ... getters and setters  
}

我使用以下代碼來創建JPA查詢

CriteriaBuilder cb = em.getCriteriaBuilder();  
CriteriaQuery<RegionTree> cq = cb.createQuery(RegionTree.class);  
Root<RegionTree> e = cq.from(RegionTree.class);  
Join<RegionTree, RegionTree> r = e.join("region", JoinType.LEFT);  
Predicate p1 = cb.greaterThan(e.get("lft").as(Integer.class), r.get("lft").as(Integer.class));  
Predicate p2 = cb.lessThan(e.get("rgt").as(Integer.class), r.get("rgt").as(Integer.class));  
Predicate p3 = cb.equal(e.get("statusCode"), "A");  
Predicate p4 = cb.equal(r.get("statusCode"), "A");  
Predicate p5 = cb.equal(r.get("regCodePar"), "ALL");  
cq.where(p1,p2,p3,p4,p5);  
TypedQuery<RegionTree> tq = em.createQuery(cq);  
l = tq.getResultList();`

這是我運行這段代碼時由Hibernate自動生成的查詢。

select  
regiontree0_.REG_CODE as REG1_7_,  
regiontree0_.LFT as LFT7_,  
regiontree0_.NAME as NAME7_,  
regiontree0_.REG_CODE_PAR as REG4_7_,  
regiontree0_.RGT as RGT7_,  
regiontree0_.STATUS_CODE as STATUS6_7_   
from  
REGION_TREE regiontree0_   
left outer join  
REGION_TREE regiontree1_   
on regiontree0_.REG_CODE_PAR=regiontree1_.REG_CODE   
where  
cast(regiontree0_.LFT as integer)>cast(regiontree1_.LFT as integer)   
and cast(regiontree0_.RGT as integer)<cast(regiontree1_.RGT as integer)   
and regiontree0_.STATUS_CODE=?   
and regiontree1_.STATUS_CODE=?   
and regiontree1_.REG_CODE_PAR=?  

我嘗試了很多方法，包括刪除cq.where代碼行，但生成的查詢與原始查詢不匹配。 我配置錯了嗎？

Answer 1

嘗試調用cq.select(r); 創建連接后。 如果沒有cq.select() ，則最后一次cq.from()調用的結果將用作選擇根。