我们有一个要求,我们需要根据父ID来获取完整的记录层次结构。我们尝试使用左外部联接,但是性能却一团糟,查询变得非常庞大。
Oracle版本:Oracle Database 19c企业版19.0.0.0.0
说,产品由孩子的组合组成,例如child1,child2 .. child5。我们在这里保持的关系是:
Product can be direct parent of child1, child2, child3, child4.
child1 can be direct parent of child2, child3 only.
child2 can be direct parent of child3, child4.
child3 can be direct parent of child4.
为简单起见,提供了3个表的数据。 _Imed_Par =子产品/子产品的直接父项
Table1 Table2 Table3
Product ProductName SPB P_ID SPC SP_B_ID P_ID
P101 Pname1 B201 P101 C301 P101
P102 Pname2 B202 P103 C302 B201 P101
P103 Pname3 B203 P103 C303 B202 P103
B204 P101 C304 B203 P103
C305 B202 P103
Expected Result:
P_ID SP_B_ID SP_C_ID Imed_PAR
P101 B201 C302 SPB
P101 B204 Product
P101 C301 Product
--- GETTING FIRST LEVEL ---
select A,B,C, 'PRODUCT' from PRODUCT PR
left outer join SPB B on B.PRODUCT_id=PR.id
left outer join SPC C on C.PRODUCT_id=PR.id AND C.SPB_ID IS NULL
left outer join SPD D on D.PRODUCT_id=PR.id AND D.SPB_ID IS NULL AND D.SPC_ID IS NULL
LEFT OUTER JOIN SPE E ON E.PRODUCT_ID=PR.id AND E.SPC_ID IS NULL AND E.SPD_ID IS NULL
UNION ALL
--- GETTING RECORDS OF ALL CHILDS WHOSE IMMEDIATE PARENT IS SPB ---
select A,B,C, 'SPB' from SPB B
left outer join SPC C on C.SPB_id=B.id
left outer join SPD D on D.SPB_id = B.id and D.SPC_id is null
--NO SPD JOIN HERE AS THERE IS NO DIRECT AND RELATION SHIP---
UNION ALL
SELECT A,B,C, 'SPC' FROM SPC C
left outer join SPD D on D.SPC_ID=C.id AND D.SPB_ID IS NULL
LEFT OUTER JOIN SPE E ON E.SPC_ID=C.id AND E.SPD_ID IS NULL
UNION ALL
SELECT A,B,C, 'SPD' FROM SPD D
LEFT OUTER JOIN SPE E ON E.SPD_ID=D.id AND E.SPC_ID IS NULL
答案 0 :(得分:-2)
当不知道父子关系中的级别数时,可以使用递归CTE方法来解决问题。在此处查看详细讨论
https://oracle-base.com/articles/11g/recursive-subquery-factoring-11gr2
但是在您给出的示例中,父子关系的级别似乎固定且已知。在这种情况下,您可以通过LEFT JOINS
处理它。如果您对此有疑问,分享您的实际查询可以帮助我们更好地理解挑战。