Question

我有两张桌子：

表一

NumLig - CodPro - CodDer
901 - BSFX30 - 0140I18
898 - RSFX30 - 18
6 - MFLX - U

表二

NumLig - CodCmp - DerCmp - QtdUti - PerCmp
6 - MDFIB0008 - null - 1 - 2
6 - MDQDN0009 - null - 0,24 - 1
898 - MFLX - U - 0,942 - 2
898 - MDCAM0002 - 0,05 - 0
901 - RSFX - 18 - 1 - 2,5
901 - EDEAD0005 - 0,245 - 0

我需要的是搜索BSFX30 - 0140I18的输出为：

MDFIB0008 - null - 1 - 2
MDQDN0009 - null - 0,24 - 1
MFLX - U - 0,942 - 2
MDCAM0002 - 0,05 - 0
RSFX - 18 - 1 - 2,5
EDEAD0005 - 0,245 - 0

如果我这样做：

SELECT E622SIM.CodPro, E622SIM.CodDer, E622DER.TipCpc, E622SIM.NumLig, E622DER.CodCmp, E622DER.DerCmp, E622DER.QtdUti, E622DER.PerCmp
FROM E622SIM
JOIN E622DER
ON E622SIM.NumLig = E622DER.NumLig

输出是所有数据，但我需要通过CodPro和CodDer过滤。如果我这样做：

SELECT E622SIM.CodPro, E622SIM.CodDer, E622DER.TipCpc, E622SIM.NumLig, E622DER.CodCmp, E622DER.DerCmp, E622DER.QtdUti, E622DER.PerCmp
FROM E622SIM
JOIN E622DER
ON E622SIM.NumLig = E622DER.NumLig

输出仅为：

RSFX - 18 - 1 - 2,5
EDEAD0005 - 0,245 - 0

但我需要：

MDFIB0008 - null - 1 - 2
MDQDN0009 - null - 0,24 - 1
MFLX - U - 0,942 - 2
MDCAM0002 - 0,05 - 0
RSFX - 18 - 1 - 2,5
EDEAD0005 - 0,245 - 0

这是因为RSFX是BSFX30的孩子，而MFLX是RSFX的孩子

谢谢！

Answer 1

我认为您需要查看有关Hierarchical Queries的文档要使用connect by连接您的数据，以获取根元素 - connect_by_root，要按root过滤元素，您需要使用where connect_by_root(E622SIM.CodPro) = 'BSFX30'

我尝试将您的描述编译成脚本

create table E622SIM(NumLig int, CodPro varchar2(100), CodDer varchar2(100));

insert into E622SIM values (901, 'BSFX30', '0140I18');
insert into E622SIM values (898, 'RSFX30', '18');
insert into E622SIM values (6, 'MFLX', 'U');

create table E622DER (NumLig int, CodCmp varchar2(100), DerCmp varchar2(100), QtdUti varchar2(100), PerCmp varchar2(100));
insert into E622DER values(901, 'RSFX30',18,1,'2,5');
insert into E622DER values(901, 'EDEAD0005',0,245,0);
insert into E622DER values(898, 'MFLX','U','0,942',2);
insert into E622DER values(898, 'MDCAM0002',null,'0,05',0);
insert into E622DER values(6, 'MDFIB0008', null, 1, 2);
insert into E622DER values(6, 'MDQDN0009',null,'0,24',1);

这是一个看起来像你需要的查询

select E622DER.CodCmp
      ,E622DER.DerCmp
      ,E622DER.QtdUti
      ,E622DER.PerCmp
from E622DER
  left outer join E622SIM on E622SIM.NumLig = E622DER.NumLig
  where connect_by_root(E622SIM.CodPro)  in ('BSFX30')
   connect by prior E622DER.CodCmp = E622SIM.CodPro  
order by level desc, E622DER.CodCmp asc

    CODCMP  DERCMP  QTDUTI  PERCMP
1   MDFIB0008       1   2
2   MDQDN0009       0,24    1
3   MDCAM0002       0,05    0
4   MFLX    U   0,942   2
5   EDEAD0005   0   245 0
6   RSFX30  18  1   2,5

Oracle中的递归SELECT

1 个答案: