Question

使用以下代码：

SELECT        SUM(LD_NUM) AS Expr1, LD_NUM
FROM            Bale
GROUP BY LD_NUM

returns Expr1 = 74987 and LD_NUM = 4411
returns Expr1 = 61768 and LD_NUM = 4412
returns Expr1 = 75021 and LD_NUM = 4413

等..

如果我74987/4411 = 17，这给了我每LD_NUM的计数

有没有办法返回关系（17,4411），（14,1412），（17,4413）并获得或命令'Expr1'进入前20名？

希望这可以自那以后。

Answer 1

SELECT TOP 20 SUM(LD_NUM) AS Expr1, LD_NUM, COUNT(LD_NUM) AS RecordCount
    FROM Bale
    GROUP BY LD_NUM
    ORDER BY Expr1 DESC

不确定您是否需要SUM用于任何其他目的。它可以很简单：

SELECT TOP 20 LD_NUM, COUNT(LD_NUM) AS RecordCount
    FROM Bale
    GROUP BY LD_NUM
    ORDER BY RecordCount DESC

Answer 2

您不需要进行任何计算来计算，

count(LD_NUM)

就是所需要的

Answer 3

试试这个

SELECT TOP 20 Expr1/LD_NUM,LD_NUM
FROM
(
   SELECT        SUM(LD_NUM) AS Expr1, LD_NUM
     FROM            Bale
    GROUP BY LD_NUM
) xx
ORDER BY xx.expr1

得到每个负载的最大数量＃并且只有前20个

3 个答案: