TSQL - それは

Question

存在しないときに私はここに、このいずれかを把握することはできませんレコードのデータを選択するには、私のテーブル構造です：

表の期間（YEARMONTHのVARCHAR（6）、weekOfMonth VARCHAR（1）、fullDate日）

表データ（SKUのvarchar型は、（6）、INT、YEARMONTHのVARCHAR（6）、fullDate日）ここに

を受けた期間のための私のデータです：これは私がしましたクエリ

yearMonth weekOfMonth fullDate 
------------------------------------------ 
201104  1  Apr 3 2011 12:00AM 
201104  1  Apr 4 2011 12:00AM 
201104  1  Apr 5 2011 12:00AM 
201104  1  Apr 6 2011 12:00AM 
201104  1  Apr 7 2011 12:00AM 
201104  1  Apr 8 2011 12:00AM 
201104  1  Apr 9 2011 12:00AM 
201104  2  Apr 10 2011 12:00AM 
201104  2  Apr 11 2011 12:00AM 
201104  2  Apr 12 2011 12:00AM 
201104  2  Apr 13 2011 12:00AM 
201104  2  Apr 14 2011 12:00AM 
201104  2  Apr 15 2011 12:00AM 
201104  2  Apr 16 2011 12:00AM 
201104  3  Apr 17 2011 12:00AM 
201104  3  Apr 18 2011 12:00AM 
201104  3  Apr 19 2011 12:00AM 
201104  3  Apr 20 2011 12:00AM 
201104  3  Apr 21 2011 12:00AM 
201104  3  Apr 22 2011 12:00AM 
201104  3  Apr 23 2011 12:00AM 
201104  4  Apr 24 2011 12:00AM 
201104  4  Apr 25 2011 12:00AM 
201104  4  Apr 26 2011 12:00AM 
201104  4  Apr 27 2011 12:00AM 
201104  4  Apr 28 2011 12:00AM 
201104  4  Apr 29 2011 12:00AM 
201104  4  Apr 30 2011 12:00AM 
201105  1  May 1 2011 12:00AM 
201105  1  May 2 2011 12:00AM 
201105  1  May 3 2011 12:00AM 
201105  1  May 4 2011 12:00AM 
201105  1  May 5 2011 12:00AM 
201105  1  May 6 2011 12:00AM 
201105  1  May 7 2011 12:00AM 
201105  2  May 8 2011 12:00AM 
201105  2  May 9 2011 12:00AM 
201105  2  May 10 2011 12:00AM 
201105  2  May 11 2011 12:00AM 
201105  2  May 12 2011 12:00AM 
201105  2  May 13 2011 12:00AM 
201105  2  May 14 2011 12:00AM 
201105  3  May 15 2011 12:00AM 
201105  3  May 16 2011 12:00AM 
201105  3  May 17 2011 12:00AM 
201105  3  May 18 2011 12:00AM 
201105  3  May 19 2011 12:00AM 
201105  3  May 20 2011 12:00AM 
201105  3  May 21 2011 12:00AM 
201105  4  May 22 2011 12:00AM 
201105  4  May 23 2011 12:00AM 
201105  4  May 24 2011 12:00AM 
201105  4  May 25 2011 12:00AM 
201105  4  May 26 2011 12:00AM 
201105  4  May 27 2011 12:00AM 
201105  4  May 28 2011 12:00AM 
201106  1  Jun 1 2011 12:00AM 
201106  1  Jun 2 2011 12:00AM 
201106  1  Jun 3 2011 12:00AM 
201106  1  Jun 4 2011 12:00AM 
201106  1  May 29 2011 12:00AM 
201106  1  May 30 2011 12:00AM 
201106  1  May 31 2011 12:00AM 
201106  2  Jun 5 2011 12:00AM 
201106  2  Jun 6 2011 12:00AM 
201106  2  Jun 7 2011 12:00AM 
201106  2  Jun 8 2011 12:00AM 
201106  2  Jun 9 2011 12:00AM 
201106  2  Jun 10 2011 12:00AM 
201106  2  Jun 11 2011 12:00AM 
201106  3  Jun 12 2011 12:00AM 
201106  3  Jun 13 2011 12:00AM 
201106  3  Jun 14 2011 12:00AM 
201106  3  Jun 15 2011 12:00AM 
201106  3  Jun 16 2011 12:00AM 
201106  3  Jun 17 2011 12:00AM 
201106  3  Jun 18 2011 12:00AM 
201106  4  Jun 19 2011 12:00AM 
201106  4  Jun 20 2011 12:00AM 
201106  4  Jun 21 2011 12:00AM 
201106  4  Jun 22 2011 12:00AM 
201106  4  Jun 23 2011 12:00AM 
201106  4  Jun 24 2011 12:00AM 
201106  4  Jun 25 2011 12:00AM 
201106  5  Jul 1 2011 12:00AM 
201106  5  Jul 2 2011 12:00AM 
201106  5  Jun 26 2011 12:00AM 
201106  5  Jun 27 2011 12:00AM 
201106  5  Jun 28 2011 12:00AM 
201106  5  Jun 29 2011 12:00AM 
201106  5  Jun 30 2011 12:00AM 

です試みた：

データがない場合は、毎年その月の月および週が返されません。

は、ここでは、テーブルデータのデータです：

SKU received yearMonth weekOfMonth 
------------------------------------- 
AAAA 5  201106  4 
BBBB 1  201106  2 
BBBB 1  201106  3 

何があるかどうか、私はNULLまたは受信のための0のいずれかを表示したいんでしょう：

sku received yearMonth fullDate 
--------------------------------------- 
AAAA 5  201106 Jun 23 2011 12:00AM 
BBBB 1  201106 Jun 10 2011 12:00AM 
BBBB 1  201106 Jun 15 2011 12:00AM 

は、ここで私は2つのテーブルを結合するとき、私のデータですたとえば、

SKU received yearMonth weekOfMonth 
------------------------------------- 
AAAA 0  201104  1 
AAAA 0  201104  2 
AAAA 0  201104  3 
AAAA 0  201104  4 
AAAA 0  201105  1 
AAAA 0  201105  2 
AAAA 0  201105  3 
AAAA 0  201105  4 
AAAA 0  201106  1 
AAAA 0  201106  2 
AAAA 0  201106  3 
AAAA 5  201106  4 
AAAA 0  201106  5 

BBBB 0  201104  1 
BBBB 0  201104  2 
BBBB 0  201104  3 
BBBB 0  201104  4 
BBBB 0  201105  1 
BBBB 0  201105  2 
BBBB 0  201105  3 
BBBB 0  201105  4 
BBBB 0  201106  1 
BBBB 1  201106  2 
BBBB 1  201106  3 
BBBB 0  201106  4 
BBBB 0  201106  5 

ありがとうございます。

Answer 1

SUM（）関数はNULL結果を排除しています。実行メッセージ：警告：Null値は、集計または他のSET操作によって削除されます。

UPDATE：
それは醜いですが、それは動作します。調整が必要な場合があり、専門知識のある人はそれを批評することができます。

UPDATE：
IはDISTINCTを加え、サブクエリにGROUP BYを除去します。

SELECT DISTINCT 
    subQ.SKU, 
    (
     SELECT SUM(d.Received) AS NumReceived 
     FROM data d 
      JOIN period p ON d.FullDate = p.FullDate 
     WHERE d.YearMonth = subQ.yearMonth 
      AND d.SKU = subQ.SKU 
      AND p.weekOfMonth = subQ.weekOfMonth 
     ) AS Received, 
    subQ.yearMonth, 
    subQ.weekOfMonth 
FROM (
     SELECT DISTINCT 
      d.SKU, 
      p.yearMonth, 
      p.weekOfMonth 
     FROM 
      period p, 
      data d 
    ) subQ 
ORDER BY 
    subQ.SKU 

Answer 2

これが結果を返さない場合は、お使いの期間テーブルから欠落したレコードが存在しなければならない：

SELECT DISTINCT 
    a.SKU, 
    SUM(a.received) AS Received 
    a.month, 
    p.weekOfMonth 
FROM 
    period p LEFT OUTER JOIN 
    data a ON a.fullDate = p.fullDate 
WHERE 
    p.month >= '04' AND p.month < '07' 
GROUP BY 
    a.SKU, 
    a.month, 
    p.weekOfMonth