如何使用按 XXX 分组，并正确使用聚合函数？具有最小值

Question

我正在尝试使用联接从 4 个表中获取记录。这些表格是费率、承运人、仓库和保证金。我能够从这 4 个表中获取所需的记录，但是当我使用 group by 和聚合函数 (MIN) 时，我的 min sell 列是正确的，但是，其他数据，如 carrierID、depotID 和 ratesID 是不同的.

我有 OriginType (OT) 和 DestinatioType (DT)，每个都有 2 个案例；仓库和门。因此，当我对它们进行分组时，我得到了四个选项（服务）：

• 仓库到仓库
• 仓库到门
• 仓库门口
• 门到门

我正在尝试为这些服务中的每一项获取最低（费率）作为“销售”并显示它。卖出计算如下：

1. 作为 A 的 autoRate 表有一个列 Buy、carrier、OT、DT、Origin (Value: Canberra)、Destination (Value: Melbourne) 和 Car (价值：4WD/Van)
2. 左加入运营商作为 C 和 A.carrier = C.ID
3. C.Fuellevy 列作为百分比((C.FuelLevy * A.buy) + C.FuelLevy) 作为 EQ1
4. ((EQ1 * 10%) + EQ1) 作为 EQ2
5. 每行的 EQ2 价格都有一个保证金表百分比。例如，如果 EQ2 的值为 400，那么它会在 Margin 表中查找范围（低和高分别为 350（低）和 500（高）），其百分比为 25%，因此((EQ2 * 25%)+ EQ2) 给出出售价值。

我不太清楚如何在这里上传我的数据和数据库表，所以我试着在这里解释一下我想要什么。

我构建的查询是：-

Select   Depo.*,  DL.id as DepoID,  DL.carrier as CarNo, DL.depotCity, DL.depoSuburb, min(Depo.Sell)  as sellcost , Depo.OriginType as OT From (
    Select Mar.*,  M.MarginPer, round((eq2 * M.MarginPer) + eq2)as Sell   From (
                        Select GST.* , EQ1 as 'FinalEQ1' , round((EQ1 * .10) + EQ1,2) as eq2 From (
                                Select A.ID as RateID, A.Origin, A.OriginState, A.Destination, A.DestinationState, A.Carrier as RateCarrier, A.Car as CarType, A.Buy as Buy, A.OriginType, A.DestinationType ,
                                C.ID as CarrierID, C.Carrier, C.FuelLevy , round((A.buy * C.FuelLevy) + A.Buy, 2) As EQ1,
                                CONCAT(A.OriginType, ' to ' ,A.DestinationType ) as service,
                                 D.id as DepoID,  D.carrier as CarNo, D.depotCity, D.depoSuburb
                                from carrier C
                                left join autorates A on A.carrier = C.ID
                                left join dList D on D.carrier =  C.ID
                                where A.origin = 'Canberra' and A.destination = 'Melbourne' and A.car = '4WD/Van'  AND  D.carrier = A.carrier AND A.goodsAllowed =  0
                                AND C.Disabled = 0
                                AND D.depotCity  = 'Canberra'
                                 order by  EQ1
                            ) As GST
                        order by  eq2
             ) As Mar
             Left Join margin M on Mar.eq2 >=  M.low and  Mar.eq2 <= M.high
                   order by  Sell
        ) As Depo
            Left  Join dList DL  on DL.Carrier = Depo.RateCarrier
            Where DL.depotCity = 'Melbourne'
             group by OT
             order by  sellcost

Group by 和 MIN() 之前的结果：-

我们在这里看到的 sellcost 值是 412，而 carNo 是 51

在此屏幕截图中，carrierno 已更改，但 MIN 值保持不变。

Answer 1

我刚刚管理了 get 查询。这将像我在我的 SQL-WB 上测试过的那样工作

SELECT DepoD.*, DL2.id, DL2.carrier AS a, DL2.depotCity
    , DL2.depoSuburb AS 'PickupSub', MIN(Sell) AS sellcost 
FROM ( 
    SELECT Depo.*, DL.id AS DepoID, DL.carrier AS CarNo, DL.depotCity, DL.depoSuburb AS 'PickupSub'
        , DL.depoSuburb AS 'DestSub', MIN(Sell) AS sellcost, Depo.OriginType AS OT 
    FROM (
        SELECT Mar.*, M.MarginPer, MIN(round((eq2 * M.MarginPer) + eq2)) AS Sell 
        FROM (
            SELECT GST.*, EQ1 AS 'FinalEQ1', MIN(round((EQ1 * .10) + EQ1, 2)) AS eq2 
            FROM (
                SELECT A.ID AS RateID, A.Origin, A.OriginState, A.Destination, A.DestinationState
                    , A.Carrier AS RateCarrier, A.Car AS CarType, A.Buy AS Buy,  A.OriginType
                    , A.DestinationType, C.ID AS CarrierID, C.Carrier, C.FuelLevy
                    , round((A.buy * C.FuelLevy) + A.Buy, 2) AS EQ1
                    , CONCAT(A.OriginType, ' to ', A.DestinationType ) AS service 
                FROM carrier C
                LEFT JOIN autorates A ON A.carrier = C.ID
                WHERE A.origin = 'Melbourne' 
                    AND A.destination = 'Canberra' 
                    AND A.car = '4WD/Van' 
                    AND A.goodsAllowed = 0
                    AND C.Disabled = 0
                ORDER BY EQ1 ASC
            ) AS GST
            GROUP BY service, CarrierID
            ORDER BY eq2
        ) AS Mar
        LEFT JOIN margin M ON Mar.eq2 >= M.low AND Mar.eq2 <= M.high
        GROUP BY service
        ORDER BY Sell
    ) AS Depo
    LEFT JOIN dList AS DL ON DL.Carrier = Depo.RateCarrier
    WHERE DL.depotCity IN('Melbourne', 'Canberra') 
    GROUP BY carrier, service
    ORDER BY sellcost
) AS DepoD
LEFT JOIN dList DL2 ON DL2.Carrier = DepoD.RateCarrier
WHERE DL2.depotCity IN('Melbourne', 'Canberra') 
GROUP BY carrier, service
ORDER BY sellcost

Answer 2

注意： 不是最终答案，而是用于测试并查看这部分是否有效的查询。如果是这样，我将提供其余的查询。

---

如果您将问题分解，我认为您需要做的是确定 OriginType 和 DestinationType 的每个组合的最低 Sell。然后加入表以获取匹配的行。

下面的查询没有得到所有的详细信息，但它应该得到具有最低 Sell 值的行。如果它有效，那么我可以添加到查询中以获取其他匹配的数据。

这个问题的一个方面是LEFT JOIN 的使用。这可能包括与ON 子句不匹配的行，本质上是添加具有空值的列。这可能会影响性能，并且仅使用 JOIN 可能会更好。如果您对此进行测试并且它的工作原理如下所述，请再次运行它，而不要在每个 JOIN 子句前面加上 LEFT。如果您得到相同的结果，请在 cmets 中告诉我。

试一试，看看您是否获得了 OriginType 和 DestinationType 的每种组合的四个值。如果是这样，我将使用查询的其余部分更新答案。

SELECT
    A.OriginType,
    A.DestinationType,
    round(A.buy * (1.0 + C.FuelLevy), 2) AS EQ1,
    round(round(A.buy * (1.0 + C.FuelLevy), 2) * 1.10, 2) AS eq2,
    MIN(round((eq2 * (1.0 + M.MarginPer)))) AS Sell
FROM carrier C
LEFT JOIN autorates A
    ON A.carrier = C.ID
LEFT JOIN dList D
    ON D.carrier = A.carrier
LEFT JOIN margin M
    ON round(round(A.buy * (1.0 + C.FuelLevy), 2) * 1.10, 2) BETWEEN M.low AND M.high
WHERE A.origin = 'Canberra'
    AND A.destination = 'Melbourne'
    AND A.car = '4WD/Van'
    AND A.goodsAllowed =  0
    AND C.Disabled = 0
    AND D.depotCity  = 'Canberra'
GROUP BY LOWER(A.OriginType), LOWER(A.DestinationType)
ORDER BY Sell