Calculating Daily Price

Posted on 2014-12-15
Last Modified: 2014-12-22
I have a table that holds price change data and I need to be able to calculate what the price was on a given date.  Example What was the price on 2014-08-05.

The ultimate goal is to get the starting and ending price for a date range.  BETWEEN 2014-08-05 AND 2014-08-20

DATE              PRICE  ITEM
2014-08-02      4.46      1
2014-08-09      4.89      1
2014-08-13      4.69      1
2014-08-25      4.99      1
2014-08-30      4.89      1
Question by:bushido121
  • 2
LVL 65

Expert Comment

by:Jim Horn
ID: 40501329
Assuming there can be only one row per id / date, give this a whirl..
SELECT id, Price, Max(Date) as current_price_date
FROM YourTable
WHERE Date <= '2014-08-05'
GROUP BY id, Price

SELECT id, Price, Max(Date) as current_price_date
FROM YourTable
WHERE Date <= '2014-08-20'
GROUP BY id, Price

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Author Comment

ID: 40501489
I don't think I explained it very well.  I'm trying to get the price for that item on 8/5 and the price on 8/20 so I can calculate the difference.

On 8/5 the item was 4.46 (I needed to go back to 8/2 to get the last price change)
On 8/20 the item was 4.69 (I needed to go back to 8/13 to get the last price change)

Ultimately I'm looking for the impact of Item 1 from 8/2 to 8/20.  (4.69 - 4.46 =  .23)

Obviously there are thousands of records in the table.

DATE              PRICE  ITEM
2014-08-02      4.46      1
2014-08-09      4.89      1
2014-08-13      4.69      1
2014-08-25      4.99      1
2014-08-30      4.89      1
LVL 69

Accepted Solution

Scott Pletcher earned 500 total points
ID: 40501611
DROP TABLE #price_data
CREATE TABLE #price_data (
    date date not null,
    price decimal(9, 2) not null,
    item varchar(10) not null
INSERT INTO #price_data SELECT '2014-08-02',      4.46,      1
INSERT INTO #price_data SELECT '2014-08-09' ,    4.89  ,    1
INSERT INTO #price_data SELECT '2014-08-13'  ,  4.69    ,  1
INSERT INTO #price_data SELECT '2014-08-25'  , 4.99  ,    1
INSERT INTO #price_data SELECT '2014-08-30'  ,4.89    ,  1

DECLARE @date1 date
DECLARE @date2 date

SET @date1 = '20140805'
SET @date2 = '20140820'

SELECT pd.item,
    MAX(CASE WHEN = item_dates.item_date1 THEN price END) AS item_price1,
    MAX(CASE WHEN = item_dates.item_date2 THEN price END) AS item_price2
    MAX(CASE WHEN = item_dates.item_date2THEN price END) -
    MAX(CASE WHEN = item_dates.item_date1THEN price END) AS item_price_diff
    SELECT pd2.item,
        MAX(CASE WHEN <= @date1 THEN END) AS item_date1,
        MAX(CASE WHEN <= @date2 THEN END) AS item_date2
    FROM #price_data pd2
    WHERE <= @date2
    GROUP BY pd2.item
) AS item_dates
INNER JOIN #price_data pd ON
    pd.item = item_dates.item AND IN (item_dates.item_date1, item_dates.item_date2)
GROUP BY pd.item
ORDER BY pd.item

Author Closing Comment

ID: 40512919
Awesome!  Thank You

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