Question

我有三个SELECT语句在多个结果集中返回：

SELECT COUNT(*) AS OrdersInStep1 FROM
(SELECT DISTINCT        
os.db_OrderNo
FROM tblOrders os
JOIN tblRefOrderType ot on ot.db_OrderType =  os.db_OrderType
LEFT JOIN tblDate d on d.db_SeqNo = os.db_Status
WHERE 
    os.db_Deleted = 0 
    AND db_Date39 = '1900/01/01' 
    AND d.db_Task = 1) AS OrdersInStep1

SELECT COUNT(*) AS OrdersInStep2 FROM
(SELECT DISTINCT        
os.db_OrderNo
FROM tblOrders os
JOIN tblRefOrderType ot on ot.db_OrderType =  os.db_OrderType   
LEFT JOIN tblDate d on d.db_SeqNo = os.db_Status
WHERE 
    os.db_Deleted = 0 
    AND db_Date39 = '1900/01/01' 
    AND d.db_Task = 2) AS OrdersInStep2

SELECT COUNT(*) AS OrdersInStep3 FROM
(SELECT DISTINCT        
os.db_OrderNo
FROM tblOrders os
JOIN tblRefOrderType ot on ot.db_OrderType =  os.db_OrderType
LEFT JOIN tblDate d on d.db_SeqNo = os.db_Status
WHERE 
    os.db_Deleted = 0 
    AND db_Date39 = '1900/01/01' 
    AND (d.db_Task = 3 OR d.db_Task = 4)
    AND d.db_SeqNo != 34 
    AND d.db_SeqNo != 31) AS OrdersInStep3

我可以使用UNION将这些结果集合并到一个列中，但是如何将它们组合成一个结果集，但是三列呢？

示例：

OrdersInStep1 | OrdersInStep2 | OrdersInStep3
     12       |      16       |      3

Answer 1

您可以构建一个具有您需要的订单的CTE，这样您就不必继续运行相同的查询3次..然后只需从CTE中选择您的计数。临时表也可以在这里接受，而不是CTE，甚至可能表现稍好......

WITH cte AS (
    SELECT 
        os.db_OrderNo,
        d.db_SeqNo,
        d.db_Task
    FROM
        tblOrders os
        JOIN tblRefOrderType ot ON ot.db_OrderType = os.db_OrderType
        JOIN tblDate d ON d.db_SeqNo = os.db_Status
    WHERE
        os.db_Deleted = 0
        AND db_Date39 = '1900/01/01'
        AND d.db_Task IN (1, 2, 3, 4)
)
SELECT
    (SELECT COUNT(DISTINCT db_OrderNo) 
      FROM cte 
      WHERE db_Task = 1) OrdersInStep1,
    (SELECT COUNT(DISTINCT db_OrderNo) 
      FROM cte 
      WHERE db_Task = 2) OrdersInStep2,
    (SELECT COUNT(DISTINCT db_OrderNo) 
      FROM cte 
      WHERE db_Task IN (3, 4) 
        AND db_SeqNo NOT IN (31, 34)) OrdersInStep3

Answer 2

你几乎就在那里，你只需要在每个select语句中都有匹配的列数。您可以根据需要传递空的''或NULL值。

SELECT COUNT(*) AS OrdersInStep1, NULL as OrdersInStep2, NULL as OrdersInStep3 FROM
(SELECT DISTINCT        
os.db_OrderNo
FROM tblOrders os
JOIN tblRefOrderType ot on ot.db_OrderType =  os.db_OrderType
LEFT JOIN tblDate d on d.db_SeqNo = os.db_Status
WHERE 
    os.db_Deleted = 0 
    AND db_Date39 = '1900/01/01' 
    AND d.db_Task = 1) AS OrdersInStep1

UNION

SELECT NULL as OrdersInStep1, COUNT(*) AS OrdersInStep2, NULL as OrdersInStep3 FROM
(SELECT DISTINCT        
os.db_OrderNo
FROM tblOrders os
JOIN tblRefOrderType ot on ot.db_OrderType =  os.db_OrderType   
LEFT JOIN tblDate d on d.db_SeqNo = os.db_Status
WHERE 
    os.db_Deleted = 0 
    AND db_Date39 = '1900/01/01' 
    AND d.db_Task = 2) AS OrdersInStep2

UNION

SELECT NULL as OrdersInStep1, NULL as OrdersInStep2, COUNT(*) AS OrdersInStep3 FROM
(SELECT DISTINCT        
os.db_OrderNo
FROM tblOrders os
JOIN tblRefOrderType ot on ot.db_OrderType =  os.db_OrderType
LEFT JOIN tblDate d on d.db_SeqNo = os.db_Status
WHERE 
    os.db_Deleted = 0 
    AND db_Date39 = '1900/01/01' 
    AND (d.db_Task = 3 OR d.db_Task = 4)
    AND d.db_SeqNo != 34 
    AND d.db_SeqNo != 31) AS OrdersInStep3

UNION将多个结果集分成不同的列

2 个答案: