Question

我正在努力使每年的月销售额排第二位。

到目前为止，我只获得了第一年第二高的月销售额。

WITH newtable AS
(
    SELECT 
    MONTH(o.orderdate) AS 'MONTH',
    YEAR(o.orderdate) AS 'YEAR',
    SUM(od.qty*od.unitprice) AS monthSales
    FROM Sales.Orders AS o
    INNER JOIN Sales.OrderDetails AS od
    ON o.orderid = od.orderid
    GROUP BY YEAR(o.orderdate), MONTH(o.orderdate) 
)
SELECT YEAR, MAX(monthSales) AS secondHighestMonthlySales
FROM newtable
WHERE monthSales < (SELECT MAX(monthSales) FROM newtable)
GROUP BY YEAR;

我每年需要第二高的人。

Answer 1

假设您在newtable中有正确的数据，请集中在有关所需内容的第二个查询上。这是纯SQL：

测试数据：

lbl2.setContentHuggingPriority(.defaultHigh, for: .horizontal)

查询以选择倒数第二个：

Year    Sales
2010    500
2010    400
2010    600
2011    700
2011    800
2011    900
2012    400
2012    600
2012    500

输出：

select O.year, max(O.sales) as secondhighestsale from Orders O,
(select year, max(sales) as maxsale
from Orders
group by year) A
where O. year = A.year
and O.sales < A.maxsale
group by O.year

Answer 2

或者，您可以使用ROWNUMBER（）函数。在这种情况下，我使用公用表表达式。我们发现每年的总销售额排名第二。这些年就是所谓的分区。

USE TEMPDB

CREATE TABLE #Sales (SalesYear INT, TotalAmount INT)
INSERT INTO #Sales VALUES (2016, 1100), (2016, 700), (2016, 950),
                          (2017, 660), (2017, 760), (2017, 460),
                          (2017, 141), (2018, 999), (2018, 499);

WITH CTE AS 

(
    SELECT *, ROW_NUMBER() OVER (PARTITION BY SalesYear ORDER BY TotalAmount DESC) AS RowNumb 
    FROM #Sales
)

SELECT SalesYear,
      TotalAmount
FROM CTE WHERE RowNumb = 2

如何获得每年第二高的月度销售额

2 个答案: