Question

我目前有一个带4个表的数据库：

客户

促销（CustID＆amp; RoomID fk）

房间（ManID fk）

管理器

我试图以百分比的形式获得客户所在的城市。但是，我只想向使用该经理进行购买的客户展示。目前，这一数额低于或超过100％。

当此查询与customer表隔离时，结果正确地加起来为100％，但是需要使用manager username作为参数从sales中获取。这就是我目前所拥有的：

SELECT
(COUNT(c.city) / (SELECT Count(CustID) FROM Customer) * 100) AS percent,     c.city AS City
FROM sale s
INNER JOIN customer c ON s.CustID = c.CustID
INNER JOIN room r ON s.RoomID = r.RoomID
INNER JOIN manager m ON r.ManID = m.ManID
WHERE m.UserName = 'manager123'
GROUP BY c.City;

修改

我想补充一点，CustID可能会在sales表中多次出现，另外，它可能为null。不确定在计数时是否需要使用distinct。把我的问题与此分开后，我觉得有必要提一下，因为我仍然无法从结果和不完整的数字中排除这些（总计<100％仍在显示）

Answer 1

你走了。

SELECT NoCity,COUNT(*)/percent *100 as percentOfCustomerBasedonManager,City
FROM (
SELECT m.UserName,COUNT(c.city) AS NoCity, COUNT(c.CustID) AS percent,c.city AS City FROM
sale s
INNER JOIN customer c ON s.CustID = c.CustID
INNER JOIN room r ON s.RoomID = r.RoomID
INNER JOIN manager m ON r.ManID = m.ManID

GROUP BY c.City ) cityTable WHERE cityTable.UserName = 'manager123'

Answer 2

这可能有所帮助：

SELECT city, ((city_count)/(customer_count)*100) as Percent
(
    SELECT city, COUNT(*) as city_count, customer_count
    FROM
    (
        SELECT distinct c.city
        FROM sale s
        INNER JOIN customer c ON s.CustID = c.CustID
        INNER JOIN room r ON s.RoomID = r.RoomID
        INNER JOIN manager m ON r.ManID = m.ManID
        WHERE m.UserName = 'manager123'
        GROUP BY c.city
    ) as t1
    LEFT JOIN
    (
        SELECT city as city2, COUNT(*) as customer_count
        FROM
        (
            SELECT c.city, distinct c.CustID
            FROM sale s
            INNER JOIN customer c ON s.CustID = c.CustID
            INNER JOIN room r ON s.RoomID = r.RoomID
            INNER JOIN manager m ON r.ManID = m.ManID
            WHERE m.UserName = 'manager123'
            GROUP BY c.city
        ) as t2
    ) as t3 on t1.city=t3.city2
) as master
GROUP BY city

Answer 3

好的，这是我遇到的解决方案，非常感谢Lim Neo指出了我正确的方向。

SELECT
round(COUNT(c.city) / 
(
    SELECT Count(b.CustID)
    FROM sale s
    INNER JOIN customer c ON s.CustID = c.CustID
    INNER JOIN room r ON s.RoomID = r.RoomID
    INNER JOIN manager m ON r.ManID = m.ManID
    WHERE m.UserName = 'Manager123'
) * 100)
AS percent, c.city AS City
FROM sale s
INNER JOIN customer c ON s.CustID = c.CustID
INNER JOIN room r ON s.RoomID = r.RoomID
INNER JOIN manager m ON r.ManID = m.ManID
WHERE m.UserName = 'Manager123'
GROUP BY c.City;

使用外键时获得结果百分比

3 个答案: