Question

我有一个SQL Server 2008 R2表的子集，如下所示：

cust_id | prod_id | day | price
--------+---------+-----+-------
137656    194528   42373   9.11
137656    194528   42374   9.11
137656    194528   42375   9.61
137656    194528   42376   9.61
137656    194528   42377   9.11
137656    194528   42378   9.11

我需要像这样对不同的价格期间进行排名：

cust_id | prod_id | day | price | rank
--------+---------+-----+-------+------
137656    194528   42373   9.11     1
137656    194528   42374   9.11     1
137656    194528   42375   9.61     2
137656    194528   42376   9.61     2
137656    194528   42377   9.11     3
137656    194528   42378   9.11     3

以便按cust_id，prod_id和day进行排序，但在价格变化时会增加排名。我试图像这样使用DENSE_RANK()：

SELECT 
    cust_id, prod_id, [day], price, 
    DENSE_RANK() OVER (ORDER BY cust_id, prod_id, price)
FROM
    @prices

返回类似于：

的内容

cust_id | prod_id | day | price | rank
--------+---------+-----+-------+------
137656    194528   42373   9.11     1
137656    194528   42374   9.11     1
137656    194528   42377   9.11     1
137656    194528   42378   9.11     1
137656    194528   42375   9.61     2
137656    194528   42376   9.61     2

显然，从排序中排除这一天会给我这些结果，但每当我按照DENSE_RANK()部分的顺序列出日期时 - 它只是将每个新的一天划分为新的ID ....

有没有人对如何运作有任何想法？感谢任何建议，如果需要可以提供更多信息

Answer 1

LAG和SUM

的第一个变体

SELECT
  *,
  1+SUM(IncCount)OVER(PARTITION BY cust_id ORDER BY [day]) [rank]
  --1+SUM(IncCount)OVER(PARTITION BY cust_id ORDER BY [day] ROWS BETWEEN unbounded preceding AND current row) [rank]
FROM
  (
    SELECT
      *,
      IIF(LAG(price)OVER(PARTITION BY cust_id ORDER BY [day])<>price,1,0) IncCount
      --CASE WHEN LAG(price)OVER(PARTITION BY cust_id ORDER BY [day])<>price THEN 1 ELSE 0 END IncCount
    FROM Test
  ) q

没有LAG

的第二个变体

WITH numCTE AS(
  SELECT *,ROW_NUMBER()OVER(PARTITION BY cust_id ORDER BY [day]) RowNum
  FROM Test
)
SELECT
  t1.*,
  1+SUM(CASE WHEN t2.price<>t1.price THEN 1 ELSE 0 END)OVER(PARTITION BY t1.cust_id ORDER BY t1.[day]) [rank]
  --1+SUM(CASE WHEN t2.price<>t1.price THEN 1 ELSE 0 END)OVER(PARTITION BY t1.cust_id ORDER BY t1.[day] ROWS BETWEEN unbounded preceding AND current row) [rank]
FROM numCTE t1
LEFT JOIN numCTE t2 ON t2.RowNum+1=t1.RowNum AND t2.cust_id=t1.cust_id

具有递归CTE的第三个变体

WITH numCTE AS(
  SELECT *,ROW_NUMBER()OVER(PARTITION BY cust_id ORDER BY [day]) RowNum
  FROM Test
),
rankCTE AS(
  SELECT RowNum,cust_id,prod_id,[day],price,1 [rank]
  FROM numCTE
  WHERE RowNum=1

  UNION ALL

  SELECT
    n.RowNum,n.cust_id,n.prod_id,n.[day],n.price,
    r.[rank]+CASE WHEN n.price<>r.price THEN 1 ELSE 0 END [rank]
  FROM numCTE n
  JOIN rankCTE r ON n.RowNum=r.RowNum+1 AND n.cust_id=r.cust_id
)
SELECT *
FROM rankCTE
OPTION(MAXRECURSION 0)

在一列上排名表，同时在另一列上排序

1 个答案: