SQL - Count number of changes in an ordered list - sql

Say I've got a table with two columns (date and price). If I select over a range of dates, then is there a way to count the number of price changes over time?
For instance:
Date | Price
22-Oct-11 | 3.20
23-Oct-11 | 3.40
24-Oct-11 | 3.40
25-Oct-11 | 3.50
26-Oct-11 | 3.40
27-Oct-11 | 3.20
28-Oct-11 | 3.20
In this case, I would like it to return a count of 4 price changes.
Thanks in advance.

You can use the analytic functions LEAD and LAG to access to prior and next row of a result set and then use that to see if there are changes.
SQL> ed
Wrote file afiedt.buf
1 with t as (
2 select date '2011-10-22' dt, 3.2 price from dual union all
3 select date '2011-10-23', 3.4 from dual union all
4 select date '2011-10-24', 3.4 from dual union all
5 select date '2011-10-25', 3.5 from dual union all
6 select date '2011-10-26', 3.4 from dual union all
7 select date '2011-10-27', 3.2 from dual union all
8 select date '2011-10-28', 3.2 from dual
9 )
10 select sum(is_change)
11 from (
12 select dt,
13 price,
14 lag(price) over (order by dt) prior_price,
15 (case when lag(price) over (order by dt) != price
16 then 1
17 else 0
18 end) is_change
19* from t)
SQL> /
SUM(IS_CHANGE)
--------------
4

Try this
select count(*)
from
(select date,price from table where date between X and Y
group by date,price )

Depending on the Oracle version use either analytical functions (see answer from Justin Cave) or this
SELECT
SUM (CASE WHEN PREVPRICE != PRICE THEN 1 ELSE 0 END) CNTCHANGES
FROM
(
SELECT
C.DATE,
C.PRICE,
MAX ( D.PRICE ) PREVPRICE
FROM
(
SELECT
A.Date,
A.Price,
(SELECT MAX (B.DATE) FROM MyTable B WHERE B.DATE < A.DATE) PrevDate
FROM MyTable A
WHERE A.DATE BETWEEN YourStartDate AND YourEndDate
) C
INNER JOIN MyTable D ON D.DATE = C.PREVDATE
GROUP BY C.DATE, C.PRICE
)

Related

SQL: How to create a daily view based on different time intervals using SQL logic?

Here is an example:
Id|price|Date
1|2|2022-05-21
1|3|2022-06-15
1|2.5|2022-06-19
Needs to look like this:
Id|Date|price
1|2022-05-21|2
1|2022-05-22|2
1|2022-05-23|2
...
1|2022-06-15|3
1|2022-06-16|3
1|2022-06-17|3
1|2022-06-18|3
1|2022-06-19|2.5
1|2022-06-20|2.5
...
Until today
1|2022-08-30|2.5
I tried using the lag(price) over (partition by id order by date)
But i can't get it right.
I'm not familiar with Azure, but it looks like you need to use a calendar table, or generate missing dates using a recursive CTE.
To get started with a recursive CTE, you can generate line numbers for each id (assuming multiple id values) in the source data ordered by date. These rows with row number equal to 1 (with the minimum date value for the corresponding id) will be used as the starting point for the recursion. Then you can use the DATEADD function to generate the row for the next day. To use the price values ​​from the original data, you can use a subquery to get the price for this new date, and if there is no such value (no row for this date), use the previous price value from CTE (use the COALESCE function for this).
For SQL Server query can look like this
WITH cte AS (
SELECT
id,
date,
price
FROM (
SELECT
*,
ROW_NUMBER() OVER (PARTITION BY id ORDER BY date) AS rn
FROM tbl
) t
WHERE rn = 1
UNION ALL
SELECT
cte.id,
DATEADD(d, 1, cte.date),
COALESCE(
(SELECT tbl.price
FROM tbl
WHERE tbl.id = cte.id AND tbl.date = DATEADD(d, 1, cte.date)),
cte.price
)
FROM cte
WHERE DATEADD(d, 1, cte.date) <= GETDATE()
)
SELECT * FROM cte
ORDER BY id, date
OPTION (MAXRECURSION 0)
Note that I added OPTION (MAXRECURSION 0) to make the recursion run through all the steps, since the default value is 100, this is not enough to complete the recursion.
db<>fiddle here
The same approach for MySQL (you need MySQL of version 8.0 to use CTE)
WITH RECURSIVE cte AS (
SELECT
id,
date,
price
FROM (
SELECT
*,
ROW_NUMBER() OVER (PARTITION BY id ORDER BY date) AS rn
FROM tbl
) t
WHERE rn = 1
UNION ALL
SELECT
cte.id,
DATE_ADD(cte.date, interval 1 day),
COALESCE(
(SELECT tbl.price
FROM tbl
WHERE tbl.id = cte.id AND tbl.date = DATE_ADD(cte.date, interval 1 day)),
cte.price
)
FROM cte
WHERE DATE_ADD(cte.date, interval 1 day) <= NOW()
)
SELECT * FROM cte
ORDER BY id, date
db<>fiddle here
Both queries produces the same results, the only difference is the use of the engine's specific date functions.
For MySQL versions below 8.0, you can use a calendar table since you don't have CTE support and can't generate the required date range.
Assuming there is a column in the calendar table to store date values ​​(let's call it date for simplicity) you can use the CROSS JOIN operator to generate date ranges for the id values in your table that will match existing dates. Then you can use a subquery to get the latest price value from the table which is stored for the corresponding date or before it.
So the query would be like this
SELECT
d.id,
d.date,
(SELECT
price
FROM tbl
WHERE tbl.id = d.id AND tbl.date <= d.date
ORDER BY tbl.date DESC
LIMIT 1
) price
FROM (
SELECT
t.id,
c.date
FROM calendar c
CROSS JOIN (SELECT DISTINCT id FROM tbl) t
WHERE c.date BETWEEN (
SELECT
MIN(date) min_date
FROM tbl
WHERE tbl.id = t.id
)
AND NOW()
) d
ORDER BY id, date
Using my pseudo-calendar table with date values ranging from 2022-05-20 to 2022-05-30 and source data in that range, like so
id
price
date
1
2
2022-05-21
1
3
2022-05-25
1
2.5
2022-05-28
2
10
2022-05-25
2
100
2022-05-30
the query produces following results
id
date
price
1
2022-05-21
2
1
2022-05-22
2
1
2022-05-23
2
1
2022-05-24
2
1
2022-05-25
3
1
2022-05-26
3
1
2022-05-27
3
1
2022-05-28
2.5
1
2022-05-29
2.5
1
2022-05-30
2.5
2
2022-05-25
10
2
2022-05-26
10
2
2022-05-27
10
2
2022-05-28
10
2
2022-05-29
10
2
2022-05-30
100
db<>fiddle here

SQL : create intermediate data from date range

I have a table as shown here:
USER
ROI
DATE
1
5
2021-11-24
1
4
2021-11-26
1
6
2021-11-29
I want to get the ROI for the dates in between the other dates, expected result will be as below
From 2021-11-24 to 2021-11-30
USER
ROI
DATE
1
5
2021-11-24
1
5
2021-11-25
1
4
2021-11-26
1
4
2021-11-27
1
4
2021-11-28
1
6
2021-11-29
1
6
2021-11-30
You may use a calendar table approach here. Create a table containing all dates and then join with it. Sans an actual table, you may use an inline CTE:
WITH dates AS (
SELECT '2021-11-24' AS dt UNION ALL
SELECT '2021-11-25' UNION ALL
SELECT '2021-11-26' UNION ALL
SELECT '2021-11-27' UNION ALL
SELECT '2021-11-28' UNION ALL
SELECT '2021-11-29' UNION ALL
SELECT '2021-11-30'
),
cte AS (
SELECT USER, ROI, DATE, LEAD(DATE) OVER (ORDER BY DATE) AS NEXT_DATE
FROM yourTable
)
SELECT t.USER, t.ROI, d.dt
FROM dates d
INNER JOIN cte t
ON d.dt >= t.DATE AND (d.dt < t.NEXT_DATE OR t.NEXT_DATE IS NULL)
ORDER BY d.dt;

Oracle SQL recursive adding values

I have the following data in the table
Period Total_amount R_total
01/01/20 2 2
01/02/20 5 null
01/03/20 3 null
01/04/20 8 null
01/05/20 31 null
Based on the above data I would like to have the following situation.
Period Total_amount R_total
01/01/20 2 2
01/02/20 5 3
01/03/20 3 0
01/04/20 8 8
01/05/20 31 23
Additional data
01/06/20 21 0 (previously it would be -2)
01/07/20 25 25
01/08/20 29 4
Pattern to the additional data is:
if total_amount < previous(r_total) then 0
Based on the filled data, we can spot the pattern is:
R_total = total_amount - previous(R_total)
Could you please help me out with this issue?
As Gordon Linoff suspected, it is possible to solve this problem with analytic functions. The benefit is that the query will likely be much faster. The price to pay for that benefit is that you need to do a bit of math beforehand (before ever thinking about "programming" and "computers").
A bit of elementary arithmetic shows that R_TOTAL is an alternating sum of TOTAL_AMOUNT. This can be arranged easily by using ROW_NUMBER() (to get the signs) and then an analytic SUM(), as shown below.
Table setup:
create table sample_data (period, total_amount) as
select to_date('01/01/20', 'mm/dd/rr'), 2 from dual union all
select to_date('01/02/20', 'mm/dd/rr'), 5 from dual union all
select to_date('01/03/20', 'mm/dd/rr'), 3 from dual union all
select to_date('01/04/20', 'mm/dd/rr'), 8 from dual union all
select to_date('01/05/20', 'mm/dd/rr'), 31 from dual
;
Query and result:
with
prep (period, total_amount, sgn) as (
select period, total_amount,
case mod(row_number() over (order by period), 2) when 0 then 1 else -1 end
from sample_data
)
select period, total_amount,
sgn * sum(sgn * total_amount) over (order by period) as r_total
from prep
;
PERIOD TOTAL_AMOUNT R_TOTAL
-------- ------------ ----------
01/01/20 2 2
01/02/20 5 3
01/03/20 3 0
01/04/20 8 8
01/05/20 31 23
This may be possible with window functions, but the simplest method is probably a recursive CTE:
with t as (
select t.*, row_number() over (order by period) as seqnum
from yourtable t
),
cte(period, total_amount, r_amount, seqnum) as (
select period, total_amount, r_amount, seqnum
from t
where seqnum = 1
union all
select t.period, t.total_amount, t.total_amount - cte.r_amount, t.seqnum
from cte join
t
on t.seqnum = cte.seqnum + 1
)
select *
from cte;
This question explicitly talks about "recursively" adding values. If you want to solve this using another mechanism, you might explain the logic in detail and ask if there is a non-recursive CTE solution.

Fill the data on the missing date range

I have a table will the data with exist data below:
Select Date, [Closing Balance] from StockClosing
Date | Closing Quantity
---------------------------
20200828 | 5
20200901 | 10
20200902 | 8
20200904 | 15
20200905 | 18
There are some missing date on the table, example 20200829 to 20200831 and 20200903.
Those closing quantity of the missing date will be follow as per previous day closing quantity.
I would like select the table result in a full range of date (show everyday) with the closing quantity. Expected result,
Date | Closing Quantity
---------------------------
20200828 | 5
20200829 | 5
20200830 | 5
20200831 | 5
20200901 | 10
20200902 | 8
20200903 | 8
20200904 | 15
20200905 | 18
Beside using cursor/for loop to insert the missing date and data 1 by 1, is that any SQL command can do it at once?
You have option to use recursive CTE.
For reference Click Here
;with cte as(
select max(date) date from YourTable
),cte1 as (
select min(date) date from YourTable
union all
select dateadd(day,1,cte1.date) date from cte1 where date<(select date from cte)
)select c.date,isnull(y.[Closing Quantity],
(select top 1 a.[Closing Quantity] from YourTable a where c.date>a.date order by a.date desc) )
as [Closing Quantity]
from cte1 c left join YourTable y on c.date=y.date
The easiest way to do this is to use LAST_VALUE along with the IGNORE NULLS option. Sadly, SQL Server does not support this. There is a workaround using analytic functions, but I would actually offer this simple option, which uses a correlated subquery to fill in the missing values:
WITH dates AS (
SELECT '20200828' AS Date UNION ALL
SELECT '20200829' UNION ALL
SELECT '20200830' UNION ALL
SELECT '20200831' UNION ALL
SELECT '20200901' UNION ALL
SELECT '20200902' UNION ALL
SELECT '20200903' UNION ALL
SELECT '20200904' UNION ALL
SELECT '20200905'
)
SELECT
d.Date,
(SELECT TOP 1 t2.closing FROM StockClosing t2
WHERE t2.Date <= d.Date AND t2.closing IS NOT NULL
ORDER BY t2.Date DESC) AS closing
FROM dates d
LEFT JOIN StockClosing t1
ON d.Date = t1.Date;
Demo

Oracle Running Subtraction

I have the below data. I want to subtract the first row from Total Qty (80) and then subtract the rest of the rows from QTY from the previous row of QTY1.
QTY QTY1 DATE TOTAL QTY
2 78 01-JAN-20 80
1 77 15-JAN-20
46 31 22-JAN-20
16 15 27-JAN-20
Is there a way to do this? Any help is greatly appreciated. Thanks
select
t.*
,first_value(TOTAL_QTY)over(order by DT) - sum(QTY)over(order by DT) as QTY1
from t;
Full example with your sample data:
with T(QTY, DT, TOTAL_QTY) as (
select 2 , to_date('01-JAN-20','dd-mon-yy'),80 from dual union all
select 1 , to_date('15-JAN-20','dd-mon-yy'),null from dual union all
select 46, to_date('22-JAN-20','dd-mon-yy'),null from dual union all
select 16, to_date('27-JAN-20','dd-mon-yy'),null from dual
)
select
t.*
,first_value(TOTAL_QTY)over(order by DT) - sum(QTY)over(order by DT) as QTY1
from t;
Result:
QTY DT TOTAL_QTY QTY1
2 2020-01-01 80 78
1 2020-01-15 77
46 2020-01-22 31
16 2020-01-27 15
SQL tables represent unordered sets. Your question seems to rely on the ordering of the rows. Let me assume you have a column that represents the ordering.
Use a cumulative sum:
select t.*,
sum(total_qty) over () - sum(qty) over (order by <ordering col>) as qty1
from t;
Here is a db<>fiddle.
Something like this (the CTE is just your data): if you add any more stuff later (in the total_qty column), then that would also get added to the total_qty calcuation (as would be typical for additions to, and subtractions from, inventory.
with d as
(select 2 qty, 78 qty1 , to_date('01-JAN-20','dd-mon-rr') datecol, 80 total_qty from dual union all
select 1, 77, to_date('15-JAN-20','dd-mon-rr'),null from dual union all
select 46 , 31, to_date('22-JAN-20','dd-mon-rr'),null from dual union all
select 16 , 15 , to_date('27-JAN-20','dd-mon-rr'),null from dual
)
select sum(total_qty) over (order by datecol) - sum(qty) over (order by datecol)
from d
You can do:
select
qty,
first_value(total_qty) over(order by date)
- sum(qty) over(order by date) as qty1,
date, total_qty
from t
order by date