0

集計関数「SUM」を使用して次のクエリを作成しました。

SELECT 
       D.PODetailID, 
       SUM(D.AcceptedQty) 
FROM STR_MRVDetail D 
INNER JOIN STR_MRVHeader H ON H.MRVID = D.MRVID 
INNER JOIN PUR_POHeader PH ON PH.POID = H.POID 
INNER JOIN PUR_PODetail PD ON PD.PODetailID = D.PODetailID
WHERE H.StatusID = 4
AND PH.POID = 839
AND (SELECT 
           SUM(AcceptedQty) 
     FROM 
           STR_MRVDetail 
     WHERE 
           STR_MRVDetail.PODetailID = PD.PODetailID) = 
                              (SELECT POQuantity FROM PUR_PODetail 
                               WHERE PUR_PODetail.PODetailID = PD.PODetailID)
                               GROUP BY D.PODetailID

現在、このクエリは 2 行を返します。このクエリのカウントを取得したい。上記のクエリの行を数える方法は?

4

2 に答える 2

2

クエリに追加@@ROWCOUNTすると、最終的な結果セットと、結果セット内のすべての行の数も得られます。

SELECT D.PODetailID, SUM(D.AcceptedQty), @@ROWCOUNT FROM STR_MRVDetail D  
INNER JOIN STR_MRVHeader H ON H.MRVID = D.MRVID  
INNER JOIN PUR_POHeader PH ON PH.POID = H.POID  
INNER JOIN PUR_PODetail PD ON PD.PODetailID = D.PODetailID 
WHERE H.StatusID = 4 
AND PH.POID = 839 
AND (SELECT SUM(AcceptedQty) FROM STR_MRVDetail  
     WHERE STR_MRVDetail.PODetailID = PD.PODetailID) =  
     (SELECT POQuantity FROM PUR_PODetail  
      WHERE PUR_PODetail.PODetailID = PD.PODetailID) 
GROUP BY D.PODetailID
于 2012-05-10T11:52:46.400 に答える
1 
SELECT COUNT(*) as Total_Rows
FROM
(
SELECT D.PODetailID as PODetailID, SUM(D.AcceptedQty) as Total_AcceptedQty
FROM STR_MRVDetail D 
INNER JOIN STR_MRVHeader H ON H.MRVID = D.MRVID 
INNER JOIN PUR_POHeader PH ON PH.POID = H.POID 
INNER JOIN PUR_PODetail PD ON PD.PODetailID = D.PODetailID
WHERE H.StatusID = 4
AND PH.POID = 839
AND (SELECT SUM(AcceptedQty) FROM STR_MRVDetail 
     WHERE STR_MRVDetail.PODetailID = PD.PODetailID) = 
     (SELECT POQuantity FROM PUR_PODetail 
      WHERE PUR_PODetail.PODetailID = PD.PODetailID)
GROUP BY D.PODetailID
) as t
于 2012-05-10T11:40:09.820 に答える