Microsoft SQL Server : getting highest cost for last purchase date - sql

Question

It's been a while since I used SQL so I'm a bit rusty. Let's say you want to compare the cost of things purchased from the previous month to this month. So an example would be a data table like this...
An item purchased on October cost $3 but the same item cost in September was $2 and $1. So you'd get the max cost of the max date (which would then be the $2 not $1). This would happen for every row of data.
I've done this with a stored scalar-value function, but when handling 100K+ rows of data, speeds are no where near fast. How would you do this with a select query in itself? What I did before was select both the max's in a select statement and only return 1, then call that function in a select statement. I want to do the same without stored procedures or functions for speed reasons. I know the following query won't work because you can only return 1 value, but it's something that I'm going for.
Select
Purchase, Item, USD,
(select MAX(Purchase), MAX(USD) from Table
where Item = 845 and MONTH(Purchase) = MONTH(Purchase) -1) LastCost
from Table
An example of what it should display can be portrayed as this.
What would be the best way to approach this?

Answer 1

Attention:
Select MAX(Purchase), MAX(USD) from Table will not return the highest cost for the highest date, but will return the highest date and the highest cost (no matter of what date).
This is how I would do this (on at least SQL Server 2012):
To get only one record per month and item (with the highest cost on the latest date), I use a numbering for the purchase date and cost (per item and month) with a descending sort order, first by date, then by cost. In the next step, I filter out only those records where the numbering is 1 (max cost for max date per item and month) and use the LAG function to access the previous cost:
WITH
numbering (Purchase, Item, Cost, p_no) AS (
SELECT Purchase,Item, Cost
,ROW_NUMBER() OVER (PARTITION BY Item, EOMONTH(Purchase) ORDER BY Purchase DESC, Cost DESC)
FROM tbl
)
SELECT Purchase, Item, Cost
, LAG(Cost) OVER (PARTITION BY Item ORDER BY Purchase) AS LastCost
FROM numbering
WHERE p_no = 1

Answer 2

SELECT Date, item, usd,
LAG(Date, 1) OVER(Order by date asc) as FormerDate,
LAG(usd, 1) OVER(Order by date asc) as FormerUsd
from (select date, item, max(usd) as usd from Data group by date, item) t
This basically returns the day before the current entry with its max price.

Answer 3

For SQL server 2017 below query will work for sample data
select purchase,item,
substring(usd,CHARINDEX(',',usd),len(usd)) as USD,
substring(usd,1,CHARINDEX(',',usd)) as lastcost from
(select max(purchase) as purchase,item, STRING_AGG (usd, ',') AS usd
from
(
select purchase,item,max(usd) as usd from t
group by purchase,item
) as T group by item
) T1

Answer 4

For your results, you need to use MAX() and ROW_NUMBER() with OVER(). Then partition the records by Item, Year, and Month. This will assure that the sort will be on each item, by each year, by each month. The ROW_NUMBER() will act as as simple way to put the last records at the top of the results, so you'll call row number 1 for each item to get the latest cost. After that, you use it as subquery to refine it as needed. For a start (your sample), you'll need to use CASE in order to split the USD (previous and Last cost). then you do the rest from there (simple methods).
I need to note that it's important to sort the records by year first, then month. then if you need to include the day, include it. This way you'll insure the records will be sorted correctly.
So, the query would look like something like this :
SELECT
MAX(Purchase) Purchase
, MAX(Item) Item
, MAX(CASE WHEN LastCost > USD THEN LastCost ELSE NULL END) USD
, MAX(CASE WHEN LastCost = USD THEN LastCost ELSE NULL END) LastCost
FROM (
SELECT
Purchase
, Item
, USD
, MAX(USD) OVER(PARTITION BY Item, YEAR(Purchase), MONTH(Purchase)) LastCost
, ROW_NUMBER() OVER(PARTITION BY Item, YEAR(Purchase), MONTH(Purchase) ORDER BY MONTH(Purchase)) RN
FROM Table
) D
WHERE
RN = 1

Answer 5

with data as (
select Item, eomonth(Purchase) as PurchaseMonth, max(USD) as MaxUSD
from T
group by Item, eomonth(Purchase)
)
select
PurchaseMonth, Item,
lag(MaxUSD) over (partition by Item order by PurchaseMonth) as PriorUSD
from data;