在Oracle中使用先前的垂直而不是水平的連接獲取路徑層級？

Question

嗨，我有一個SQL語句使用connect by prior。我想通過特定的where條件獲取路徑。以下SQL工作正常，但我不想只獲得匹配條件的單個行，我還想獲取每個父記錄（垂直向上層次結構）。

SELECT SUBSTR(SYS_CONNECT_BY_PATH(R.CALCULATION_PLAN_CODE, '.'),4) CODE_PATH, 
     SUBSTR(SYS_CONNECT_BY_PATH(R.CALCULATION_PLAN_NAME, ' > '),29) NAME_PATH,   
     R.CALCULATION_PLAN_CODE, 
     R.CALCULATION_PLAN_NAME   
    FROM PRM.MOVABLE_CALCULATION_PLAN R 
    WHERE R.RELATED_YEAR = :relatedYear 
    AND R.CALCULATION_PLAN_PARENT_OID != 0 
    AND REGEXP_LIKE (R.CALCULATION_PLAN_NAME,'ROL') 
    AND CONNECT_BY_ISLEAF = 1 
START WITH R.OID = 
      (SELECT R1.OID 
      FROM PRM.MOVABLE_CALCULATION_PLAN R1 
      WHERE (R1.CALCULATION_PLAN_PARENT_OID IS NULL OR R1.CALCULATION_PLAN_PARENT_OID = 0) AND R1.RELATED_YEAR=:relatedYear) CONNECT_BY_PRIOR R.OID = R.CALCULATION_PLAN_PARENT_OID; 

我不希望這樣的輸出： 我希望它是這樣的：

150  İLK MADDE VE MALZEMELER 
150.1  İLK MADDE VE MALZEMELER > KIRTASİYE MALZEMELERİ GRUBU 
150.1.1 İLK MADDE VE MALZEMELER > KIRTASİYE MALZEMELERİ GRUBU > ROL <----- 


150  İLK MADDE VE MALZEMELER 
150.1  İLK MADDE VE MALZEMELER > KIRTASİYE MALZEMELERİ GRUBU 
150.1.1 İLK MADDE VE MALZEMELER > KIRTASİYE MALZEMELERİ GRUBU > YAZI ARAÇLARI 
150.1.1.1 İLK MADDE VE MALZEMELER > KIRTASİYE MALZEMELERİ GRUBU > YAZI ARAÇLARI > ROLLER KALEM <----- 

如何實現這一目標？

在此先感謝

Answer 1

一個解決方案將需要解決你的表不止一次，因爲你需要的輸出可以有重複（如「150.1」）可能導致利率兩個不同的葉子部分路徑，所以它會發生兩次（或更多）。

請注意，您不需要重複start with子句中的where條件。無論如何也應用它們，也用於起始記錄。

下面是該查詢：

SELECT NODE.CODE_PATH, NODE.NAME_PATH FROM 
     (
     SELECT  OID, 
       SYS_CONNECT_BY_PATH(R.OID, ',') OID_PATH 
     FROM  MOVABLE_CALCULATION_PLAN R 
     WHERE  REGEXP_LIKE (R.CALCULATION_PLAN_NAME,'ROL') 
       AND CONNECT_BY_ISLEAF = 1 
       AND R.RELATED_YEAR = :relatedYear 
     START WITH NVL(CALCULATION_PLAN_PARENT_OID, 0) = 0 
     CONNECT BY PRIOR R.OID = R.CALCULATION_PLAN_PARENT_OID 
    ) LEAF 
INNER JOIN 
     (
     SELECT  SUBSTR(SYS_CONNECT_BY_PATH(R.CALCULATION_PLAN_CODE, '.'),2) CODE_PATH, 
       SUBSTR(SYS_CONNECT_BY_PATH(R.CALCULATION_PLAN_NAME, ' > '),4) NAME_PATH, 
       OID, 
       SYS_CONNECT_BY_PATH(R.OID, ',') OID_PATH 
     FROM  MOVABLE_CALCULATION_PLAN R 
     START WITH NVL(CALCULATION_PLAN_PARENT_OID, 0) = 0 
     CONNECT BY PRIOR R.OID = R.CALCULATION_PLAN_PARENT_OID 
    ) NODE 
ON NODE.OID = LEAF.OID 
OR LEAF.OID_PATH LIKE '%,' || NODE.OID || ',%' 

結果：

CODE_PATH NAME_PATH 
---------- ------------------------------------------------------------ 
150   İLK MADDE VE MALZEMELER 
150.1  İLK MADDE VE MALZEMELER > KIRTASİYE MALZEMELERİ GRUBU 
150.1.1  İLK MADDE VE MALZEMELER > KIRTASİYE MALZEMELERİ GRUBU > ROL 
150   İLK MADDE VE MALZEMELER 
150.1  İLK MADDE VE MALZEMELER > KIRTASİYE MALZEMELERİ GRUBU 
150.1.1  İLK MADDE VE MALZEMELER > KIRTASİYE MALZEMELERİ GRUBU > YAZI ARAÇLARI 
150.1.1.1 İLK MADDE VE MALZEMELER > KIRTASİYE MALZEMELERİ GRUBU > YAZI ARAÇLARI > ROL 

而一個fiddle。

想法是，在第一個子選擇中，您選擇葉子，並提供where條件。然後，你做另一個子選擇，其選擇所有節點與他們的路徑，沒有任何過濾。最後，將這些節點與第一個過濾的葉節點的下降線匹配。

Answer 2

你想要的是相當複雜的，但它是可以解決的。但是，您需要構建多個樹：通過第一個查詢，您通過以連接，以葉單位開始，構建分支到根節點，並跟蹤您正在構建分支的葉，使用connect_by_root函數。然後，在第二個查詢中，您基於第一個查詢的結果connect by，這次按照適當的順序，但是另外匹配您之前跟蹤的葉ID。

with Q as 
(
SELECT CONNECT_BY_ROOT(R.OID) leafid, R.* 
FROM MOVABLE_CALCULATION_PLAN R 
CONNECT BY R.OID = PRIOR R.CALCULATION_PLAN_PARENT_OID 
START WITH R.RELATED_YEAR = 2015 
    AND REGEXP_LIKE (R.CALCULATION_PLAN_NAME,'ROL') 
    AND R.CALCULATION_PLAN_PARENT_OID != 0 
    /*Following condition is replacement for CONNECT_BY_ISLEAF = 1, 
    Since you're connecting in reverse you can't use it*/ 
    AND not exists 
    (select 'x' from MOVABLE_CALCULATION_PLAN R2 
    where R2.CALCULATION_PLAN_PARENT_OID = R.OID) 
) 

select SUBSTR(SYS_CONNECT_BY_PATH(Q.CALCULATION_PLAN_CODE, '.'),2) CODE_PATH, 
     SUBSTR(SYS_CONNECT_BY_PATH(Q.CALCULATION_PLAN_NAME, ' > '),4) NAME_PATH 
    from Q 
    CONNECT BY PRIOR Q.OID = Q.CALCULATION_PLAN_PARENT_OID 
    and prior Q.leafid = leafid 
    START WITH NVL(CALCULATION_PLAN_PARENT_OID, 0) = 0 

像往常一樣通過查詢連接，性能可能是一個問題，但是從我對類似的表結構，看到它仍然是相當快的。另請參見http://sqlfiddle.com/#!4/dd5b68/6

P.S.在將來，一定要向我們提供樣本數據和對您的條件的恰當描述，不要讓我們破譯您的查詢並自己想想樣本。