Question

我有1个表ventory_movement，这里是表中的数据

product_id | staff_name |  status | sum | reference_number
--------------------------------------------------
  1          zes             cp     1    000122
  2          shan             cp     4    000133

我还有另一个表stock_orderproduct，其中有费用日期

 orderdate   product_id   cost
--------------------------------
01/11/2018    1            3200
01/11/2018    2             100
02/11/2018    1            4000
02/11/2018    1            500
03/11/2018    2            2000

我想要这个结果

product_id| staff_name | status | sum reference_number  |  cost
--------------------------------------------------------------
      1         zes        cp     1    000122         4000
      2         shan       cp     4    000133         2000

这是我的查询

select ipm.product_id, 
case when ipm.order_by_id is not null then 
(select au.first_name from users_staffuser us inner join auth_user au on us.user_id= au.id
where us.id = ipm.order_by_id) else '0' end as "Staff_name"
,ipm.status,
Sum(ipm.quantity), ip.reference_number
from inventory_productmovement ipm
inner join inventory_product ip on ipm.product_id = ip.id
inner join users_staffuser us on ip.branch_id = us.branch_id 
inner join auth_user au on us.user_id = au.id
AND ipm.status = 'CP'
group by ipm.product_id, au.first_name, ipm.status,  
ip.reference_number, ip.product_name
order by 1

Answer 1

您可以使用以下查询获取产品的最高费用

SELECT i.product_id,i.staff_name,i.status,i.sum reference_number ,s.MAXCost
FROM (SELECT product_id,MAX(cost) AS MAXCost
    FROM inventory_orderproduct 
    GROUP BY product_id ) s
JOIN inventory_movement i  ON i.product_id =s.product_id

要使用最新日期检索费用，请使用以下查询

WITH cte as (
   SELECT product_id,cost
         ,ROW_NUMBER() OVER (PARTITION BY product_id ORDER BY orderdate DESC) AS Rno
   FROM inventory_orderproduct )
 SELECT i.product_id,i.staff_name,i.status,i.sum reference_number ,s.Cost
 FROM cte s
 JOIN inventory_movement i  ON i.product_id =s.product_id 
 WHERE s.Rno=1

Answer 2

您可以使用下面的查询，它将根据最新日期选择数据

`[[<-.foo_list` <- function(x, i, y) {
  if (!is.foo(y)) {
    stop("Please provide an object of class foo.")
  }
  x[[i]] <- y
}

Answer 3

我认为这会对您有所帮助：SQL: Select maximum value for each unique key? 在获得每个ID的最大值的视图之后，您可以进行常规的内部联接。

Answer 4

这是您问题的解决方案，它的工作正常。如果您喜欢答案，请投票！

SELECT i.product_id,i.staff_name,i.status,i.sum reference_number ,s.Cost
FROM (SELECT product_id,MAX(cost) AS Cost
    FROM inventory_orderproduct 
    GROUP BY product_id ) s
JOIN inventory_movement i  ON i.product_id =s.product_id

Answer 5

在给定的情况下，这应该可以正常工作：

Select table1.product_id, table2.staff_name, table2.status, table2.reference_number, 
MAX(table1.cost)
FROM table2
LEFT JOIN table1 ON table1.product_id = table2.product_id
GROUP BY table2.product_id, table2.staff_name, table2.status, table2.reference_number

如何使用一个表中的最新日期并连接到另一个表来获取价值

5 个答案: