Question

我一直在尝试进行以下练习，即使用SQL创建查询，＆＃34;查找一周内拥有3个以上客户的推销员数量。显示员工ID，员工姓名，客户ID，客户名称以及销售员会见的客户数量。＆＃34;

实体和相关领域是：

Staff:   [Staff_ID (PK), Staff_Name, ...]
Clients: [Client_ID (PK), Client_Name, Staff_ID (FK), ...]
Sales:   [Sale_ID (PK), Client_ID (FK), Staff_ID (FK), Date_of_Sale, ...]
(All data are strings except Date_of_Sale, which is in DATE format)

我试图通过以下方式实现这一目标：

SELECT t.Staff_ID, t.Staff_Name, COUNT(s.Client_ID), 
COUNT(c.Client_Name), DATEPART(wk, s.Date_of_Sale) as Week
FROM Clients c, Staff t, Sales s
GROUP BY DATEPART(wk, s.Date_of_Sale), t.Staff_ID, t.Staff_Name
HAVING COUNT(DATEPART(wk, s.Date_of_Sale)) > 3

我知道它不正确，但我不知道如何修复它，因为我在sql中非常不熟练。我想了解如何做到这一点，我可以（在有限的程度上）通过逆向工程来实现。更好的是，如果有人要为我解释，我会非常感激，谢谢你。

Answer 1

我们走了

WITH CTE AS (
    SELECT *, DATEPART(wk, s.Date_of_Sale) as SALES_WEEK
    FROM Clients c
    INNER JOIN Sales s ON c.Client_ID = s.CLIENT_ID
    INNER JOIN Staff t ON t.STAFF_ID = s.STAFF_ID
)

SELECT SALES_WEEK, Staff_ID, Staff_Name, COUNT(1) AS NO_OF_SALES
FROM CTE
GROUP BY SALES_WEEK, Staff_ID, Staff_Name
HAVING COUNT(1) > 3

希望这可以按预期工作

Answer 2

推销员遇到的客户数与推销员的销售数之间存在很大差异。主要是，是否计算对同一客户的重复访问次数。前者说不，后者说是，因为客户的数量没有变化，但销售数量却有所不同。

-- Get list of clients with sales greater than 3 in a week WITH SalesCount AS ( SELECT Client_ID AS 'Client', DATEPART(wk, Date_of_Sale) AS 'week', COUNT(Staff_ID) AS 'Sales_Count' FROM Sales GROUP BY DATEPART(wk, Date_of_Sale), Client_ID HAVING COUNT(Staff_ID) > 3 ), -- Shorten the list to unique entries and add the StaffID ClientList AS ( SELECT DISTINCT sc.Client, c.Staff_ID FROM SalesCount AS sc INNER JOIN Clients AS c ON sc.Client = c.Client_ID) , -- Create a count of clients visited ClientCount AS ( SELECT COUNT(l.Client) AS visitors, l.Staff_ID FROM ClientList AS l GROUP BY l.Staff_ID) SELECT cc.visitors, cc.Staff_ID, l.Client, t.Staff_Name FROM ClientCount AS cc INNER JOIN ClientList AS l ON cc.Staff_ID = l.Staff_ID INNER JOIN Staff AS t ON t.Staff_ID = l.Staff_ID INNER JOIN Clients AS c ON l.Client = c.Client_ID

如何在查询中显示以下数据？

2 个答案: