sql与max和多个连接

643ylb08 于 2021-06-20 发布在 Mysql

关注(0)|答案(2)|浏览(358)

我试图得到每个客户在mysql上为每个产品支付的最后价格。下面的sql没有提供正确的数据。max（datelasfullfillment）不包含行值，甚至不包含max v值。这就像是最大值之前的工作组。

select 
    'item' AS type, soitem.productnum as 'SKU',
    (soitem.unitprice / right(uom.code, length(uom.code) - 2)) as unitPrice, 
    replace(customer.name, "#", "") AS priceList, 
    max(soitem.dateLastFulfillment) 
from 
    soitem
left join 
    so ON so.id = soitem.soid
left join
    customer on so.customerid = customer.id
left join 
    product on product.num = soitem.productnum
left join  
    uom on product.uomid = uom.id
where 
    soitem.dateLastFulfillment > now() - interval 6 month 
    and soitem.unitprice > 0 
    and so.statusid in (20, 25, 60)
group by 
    soitem.productnum, customer.name
order by 
    PriceList

下面是一些带有预期结果的表格示例。sql必须以select语句、NODECLARE等开头，除非没有其他选项。
so表：

id  billToName  customerid  dateCompleted  dateCreated  dateIssued  num
1  Name1  1  6/27/18  6/23/18  6/23/18  ordernum1
2  Name1  1  7/15/18  7/10/18  7/10/18  ordernum2
3  Name1  1  7/29/18  7/20/18  7/20/18  ordernum3
4  Name2  2  6/31/2018  6/30/18  6/30/18  ordernum4
5  Name2  2  7/27/18  7/26/18  7/26/18  ordernum5
6  Name3  3  8/8/18  8/5/18  8/5/18  ordernum6
7  Name3  3  7/25/18  7/20/18  7/20/18  ordernum7

soitem表格：

id  soId  unitPrice  dateLastFulfillment  productId  productNum  statusId  uomId  qtyOrdered
1  1  10  6/27/18  1  SKU-1  50  11  3
2  1  20  6/27/18  2  SKU-2  50  12  5
3  1  30  6/27/18  3  SKU-3  50  13  6
4  2  11  7/15/18  1  SKU-1  50  11  11
5  2  21  7/15/18  2  SKU-2  50  12  44
6  2  31  7/15/18  3  SKU-3  50  13  5
7  3  12  7/29/18  1  SKU-1  50  11  5
8  3  22  7/29/18  2  SKU-2  50  12  6
9  4  23  6/31/2018  2  SKU-2  50  12  9
10  4  33  6/31/2018  3  SKU-3  50  13  12
11  5  24  7/27/18  2  SKU-2  50  12  14
12  5  34  7/27/18  3  SKU-3  50  13  35
13  6  25  8/8/18  2  SKU-2  50  12  22
14  6  35  8/8/18  3  SKU-3  50  13  55
15  7  26  7/25/18  2  SKU-2  50  12  22
16  7  36  7/25/18  3  SKU-3  50  13  11

产品表：

num  uomid
SKU-1  11
SKU-2  12
SKU-3  13

计量单位表：

id  code
11  cs10
12  cs20
13  cs30

客户表：

ID  NAME
1  CUSTOMER1#
2  CUSTOMER2#
3  CUSTOMER3#

预期结果：

type  SKU  unitPrice  priceList  max(soitem.dateLastFulfillment)
item  SKU-1  1.2  customer1  7/29/18
item  SKU-2  1.1  customer1  7/29/18
item  SKU-3  1.03  customer1  7/15/18
item  SKU-2  1.2  customer2  7/27/18
item  SKU-3  1.13  customer2  7/27/18
item  SKU-2  1.25  customer3  8/8/18
item  SKU-3  1.17  customer3  8/8/18

sql mysql Join group-by Max

来源：https://stackoverflow.com/questions/51868773/sql-with-max-and-several-joins

2条答案

按热度按时间

flvtvl501#

试试这个，只是代码的第一部分，看看你得到了什么。然后粘贴其他连接等（删除前两个旧where项和所有group by）

-- compute this once, instead of for each row
Declare @now_minus_6mos as date = DATEADD(month, -6, GETDATE());
print @now_minus_6mos;

select 
    'item' AS type, soitem.productnum as 'SKU',
    (soitem.unitprice / 1 ) as unitPrice,   -- I do not have UOM, so simplify to be one
    -- I do not have Customer    replace(customer.name, "#", "") AS priceList, 
    (soitem.dateLastFulfillment) -- remove the max, since we are getting only the last one
from 
    (Select * From
        (Select
            productnum
            ,unitprice
            ,dateLastFulfillment
            ,Row_Number() Over(Partition By productnum  Order By dateLastFulfillment DESC) as dlfRow
         From soitem
         --move where filters here to reduce number of rows returned
         Where
             soitem.dateLastFulfillment > @now_minus_6mos 
         and soitem.unitprice > 0 
         ) aa
     Where dlfRow = 1
     ) soitem

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czfnxgou2#

你可以用“排名”功能来实现。它们在mysql中并不存在，但是你可以模仿它们。
请看这篇文章。
mysql中的rank函数

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