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私のソースが次のように見えると仮定します

X1  H_ID    l1_no   l2_no   l3_no
20  1       2       12      42
20  1       6       16      42
20  1       1       19      41
20  1       5       15      41

そして、私のルックアップテーブルは次のようになります

X1  H_ID    l1_no   l1_key  l2_no   l2_key  l3_no   l3_key
20  1       4       104         14      114         44      144
20  1       3       103         13      113         43      143
20  1       1       101         11      111         41      141
20  1       2       102         12      112         42      142

この場合、完全に一致し、完全なキーが保持される場合にのみ、私のレコードは更新と見なされます。どの列にも一致しない場合は、挿入と見なされます。トリッキーな部分は挿入の場合でもあり、すべての階層レベルをチェックする必要があり、一致するビットのためにキーを保持する必要があります。

たとえば、すでに 1->11->14 があり、1->11->15 を確認すると、これは挿入レコードである必要がありますが、1 と 11 のキーは保持する必要があり、15 の場合は生成する必要があります。

また、1->12->14 を取得した場合、この場合、レコードは挿入と見なされ、キーは 1 に対してのみ保持する必要があり、12 と 14 に対して新しいキーを生成する必要があります

私が考えることができる 1 つの可能な解決策は、データ セットを 2D マトリックスとして扱うことによって、各レベルの左結合を行うことです。しかし、それは素晴らしい方法ではありません。

WITH lookup ( x1, h_id,  l1_no, l1_key, l2_no, l2_key,l3_no,l3_key ) AS
         (SELECT 20, 1, 1, 101, 11, 111, 41, 141 FROM sys.dual UNION ALL
          SELECT 20, 1, 2, 102, 12, 112, 42, 142 FROM sys.dual UNION ALL
          SELECT 20, 1, 3, 103, 13, 113, 43, 143 FROM sys.dual UNION ALL
          SELECT 20, 1, 4, 104, 14, 114, 44, 144 FROM sys.dual),
               dwh_keys AS 
         (SELECT x1, h_id, MAX(l1_key) mk1,  MAX(l2_key) mk2,  MAX(l3_key) mk3 
          FROM  lookup GROUP BY 1,2) ,
     src ( x1, h_id, l1_no,  l2_no, l3_no ) AS
         (SELECT 20, 1, 1,  19, 41  FROM sys.dual UNION ALL
          SELECT 20, 1, 2,  12, 42  FROM sys.dual UNION ALL
          SELECT 20, 1, 5,  15, 41  FROM sys.dual UNION ALL
          SELECT 20, 1, 6,  16, 42  FROM sys.dual),
     stg1 AS
         (SELECT      b.*, 
                      CASE WHEN a.x1 IS NOT NULL THEN 'U' ELSE 'I' END AS delta_type
          FROM        src b LEFT OUTER JOIN lookup a ON a.x1 = b.x1 AND a.h_id = b.h_id 
          AND a.l1_no = b.l1_no AND a.l2_no = b.l2_no AND a.l3_no = b.l3_no)
SELECT      a.*, 
         CASE WHEN delta_type = 'I' AND b.l1_no IS NULL  THEN mk1 + ROW_NUMBER ( ) OVER (ORDER BY delta_type) ELSE b.l1_key END AS new1_key,
         CASE WHEN delta_type = 'I' AND c.l2_no IS NULL  THEN mk2 + ROW_NUMBER ( ) OVER (ORDER BY delta_type) ELSE c.l2_key END AS new1_key,
         CASE WHEN delta_type = 'I' AND d.l3_no IS NULL  THEN mk3 + ROW_NUMBER ( ) OVER (ORDER BY delta_type) ELSE d.l3_key END AS new1_key,
         delta_type
FROM            stg1 a 
LEFT OUTER JOIN lookup b ON a.x1 = b.x1 AND a.h_id = b.h_id AND a.l1_no = b.l1_no 
LEFT OUTER JOIN lookup c ON a.x1 = c.x1 AND a.h_id = c.h_id AND a.l1_no = c.l1_no AND a.l2_no = c.l2_no         
LEFT OUTER JOIN lookup d ON a.x1 = d.x1 AND a.h_id = d.h_id AND a.l1_no = d.l1_no AND a.l2_no = d.l2_no AND a.l3_no = d.l3_no 
LEFT OUTER JOIN dwh_keys xx ON a.x1 = xx.x1 AND a.h_id = xx.h_id;

私が得る出力（また私が欲しいもの）

X1  H_ID    L1_NO   L2_NO   L3_NO   DELTA_TYPE  NEW1_KEY    NEW1_KEY    NEW1_KEY    DELTA_TYPE
20  1   6   16  42  I   105 115 145 I
20  1   5   15  41  I   106 116 146 I
20  1   1   19  41  I   101 117 147 I
20  1   2   12  42  U   102 112 142 U

私のデータベースWX2は再帰 CTE をサポートしていません。私が持っているのは再帰 CTE を除いて単純な ANSI SQL 構文だけです。

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1 に答える 1

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私の最初の提案の修正版ですが、最終的に解決策を見つけました

WITH lookup ( x1, h_id, l1_no, l1_key, l2_no, l2_key, l3_no, l3_key ) AS
         (SELECT 20, 1, 1, 101, 11, 111, 41, 141 FROM sys.DUAL UNION ALL
          SELECT 20, 1, 1, 101, 12, 112, 42, 142 FROM sys.DUAL UNION ALL
          SELECT 20, 1, 3, 102, 13, 113, 43, 143 FROM sys.DUAL UNION ALL
          SELECT 20, 1, 4, 103, 14, 114, 44, 144 FROM sys.DUAL),
     dwh_keys AS
         (SELECT      x1, h_id, MAX ( l1_key ) mk1, MAX ( l2_key ) mk2, MAX ( l3_key ) mk3
          FROM        lookup
          GROUP BY    1, 2),
     src ( x1, h_id, l1_no, l2_no, l3_no ) AS
         (SELECT 20, 1, 1, 19, 41 FROM sys.DUAL UNION ALL
          SELECT 20, 1, 1, 12, 42 FROM sys.DUAL UNION ALL
          SELECT 20, 1, 5, 15, 41 FROM sys.DUAL UNION ALL
          SELECT 20, 1, 6, 16, 42 FROM sys.DUAL),
     stg1 AS
         (SELECT    DISTINCT b.*, CASE WHEN a.x1 IS NOT NULL THEN 'U' ELSE 'I' END AS delta_type
          FROM      src b
                    LEFT OUTER JOIN
                    lookup a
                        ON     a.x1 = b.x1
                           AND a.h_id = b.h_id
                           AND a.l1_no = b.l1_no
                           AND a.l2_no = b.l2_no
                           AND a.l3_no = b.l3_no),
     stg2 AS
         (SELECT    a.*, CASE WHEN delta_type = 'I' AND b.l1_no IS NULL THEN 0 ELSE 1 END AS flag1, b.l1_key AS k1,
                    CASE WHEN delta_type = 'I' AND c.l2_no IS NULL THEN 0 ELSE 1 END AS flag2, c.l2_key AS k2,
                    CASE WHEN delta_type = 'I' AND d.l3_no IS NULL THEN 0 ELSE 1 END AS flag3, d.l3_key AS k3
          FROM      stg1 a
                    LEFT OUTER JOIN (SELECT DISTINCT x1, h_id, l1_no, l1_key FROM lookup) b
                        ON a.x1 = b.x1 AND a.h_id = b.h_id AND a.l1_no = b.l1_no
                    LEFT OUTER JOIN (SELECT DISTINCT x1, h_id, l1_no, l1_key, l2_no, l2_key FROM lookup) c
                        ON a.x1 = c.x1 AND a.h_id = c.h_id AND a.l1_no = c.l1_no AND a.l2_no = c.l2_no
                    LEFT OUTER JOIN
                    (SELECT DISTINCT x1, h_id, l1_no, l1_key, l2_no, l2_key, l3_no, l3_key FROM lookup) d
                        ON     a.x1 = d.x1
                           AND a.h_id = d.h_id
                           AND a.l1_no = d.l1_no
                           AND a.l2_no = d.l2_no
                           AND a.l3_no = d.l3_no),
     keys AS
         (SELECT    a.*,
                    CASE WHEN flag1 = 0 THEN mk1 + DENSE_RANK ( ) OVER (ORDER BY a.delta_type, l1_no) ELSE a.k1 END AS new_key1,
                    CASE WHEN flag2 = 0 THEN mk2 + DENSE_RANK ( ) OVER (ORDER BY a.delta_type, l1_no, l2_no) ELSE a.k2 END AS new_key2,
                    CASE WHEN flag3 = 0 THEN mk3 + DENSE_RANK ( ) OVER (ORDER BY a.delta_type, l1_no, l2_no, l3_no) ELSE a.k3 END AS new_key3
          FROM      stg2 a LEFT OUTER JOIN dwh_keys b ON a.x1 = b.x1 AND a.h_id = b.h_id)
SELECT      x1, h_id, l1_no, l2_no, l3_no, delta_type, new_key1, new_key2, new_key3
FROM        keys
ORDER BY    1, 2, 3, 4, 5;
于 2016-05-09T16:44:50.147 に答える