I've the following code
declare #test table (id int, [Status] int, [Date] date)
insert into #test (Id,[Status],[Date]) VALUES
(1,1,'2018-01-01'),
(2,1,'2018-01-01'),
(1,1,'2017-11-01'),
(1,2,'2017-10-01'),
(1,1,'2017-09-01'),
(2,2,'2017-01-01'),
(1,1,'2017-08-01'),
(1,1,'2017-07-01'),
(1,1,'2017-06-01'),
(1,2,'2017-05-01'),
(1,1,'2017-04-01'),
(1,1,'2017-03-01'),
(1,1,'2017-01-01')
SELECT
id,
[Status],
MIN([Date]) OVER (PARTITION BY id,[Status] ORDER BY [Date],id,[Status] ) as WindowStart,
max([Date]) OVER (PARTITION BY id,[Status] ORDER BY [Date],id,[Status]) as WindowEnd,
COUNT(*) OVER (PARTITION BY id,[Status] ORDER BY [Date],id,[Status] ) as total
from #test
But the result is this:
id Status WindowStart WindowEnd total
1 1 2017-01-01 2017-01-01 1
1 1 2017-01-01 2017-03-01 2
1 1 2017-01-01 2017-04-01 3
1 1 2017-01-01 2017-06-01 4
1 1 2017-01-01 2017-07-01 5
1 1 2017-01-01 2017-08-01 6
1 1 2017-01-01 2017-09-01 7
1 1 2017-01-01 2017-11-01 8
1 1 2017-01-01 2018-01-01 9
1 2 2017-05-01 2017-05-01 1
1 2 2017-05-01 2017-10-01 2
2 1 2018-01-01 2018-01-01 1
2 2 2017-01-01 2017-01-01 1
And I need to be grouped by window like this.
id Status WindowStart WindowEnd total
1 1 2017-01-01 2017-04-01 3
1 2 2017-05-01 2017-05-01 1
1 1 2017-06-01 2017-09-01 4
1 2 2017-10-01 2017-10-01 1
1 1 2017-11-01 2018-01-01 2
2 1 2018-01-01 2018-01-01 1
2 2 2017-01-01 2017-01-01 1
The first group for the id= 1 Status = 1 should end at the first row with Status = 2 (2017-05-01) so the total is 3 and then start again from the 2017-06-01 to 2017-09-01 with a total of 4 rows.
How can get this done?
This is a "classic" Groups and Island issue. There's probably 1000's of answers for these on the Internet.
This works for what you're after, however, try having a bit more of a research before hand. :)
WITH Groups AS(
SELECT t.*,
ROW_NUMBER() OVER (PARTITION BY id ORDER BY [Date]) -
ROW_NUMBER() OVER (PARTITION BY id, [status] ORDER BY [Date]) AS Grp
FROM #test t)
SELECT G.id,
G.[Status],
MIN([Date]) AS WindowStart,
MAX([date]) AS WindowsEnd,
COUNT(*) AS Total
FROM Groups G
GROUP BY G.id,
G.[Status],
G.Grp
ORDER BY G.id, WindowStart;
Note, that the ordering of your last 2 lines is the other way round in this solution; it seems you're ordering ASCENDING for id 1, for DESCENDING for id 2 in your expected results.
Here is one way using LAG function
;WITH cte
AS (SELECT *,
grp = Sum(CASE WHEN prev_val = Status THEN 0 ELSE 1 END)
OVER(partition BY id ORDER BY Date)
FROM (SELECT *,
prev_val = Lag(Status)OVER(partition BY id ORDER BY Date)
FROM #test) a)
SELECT id,
Status,
WindowStart = Min(date),
WindowEnd = Max(date),
Total = Count(*)
FROM cte
GROUP BY id, Status, grp
Using lag function first find the previous status of each date, then using Sum over() create a group by incrementing the number only when there is a change in status.
Related
I have the following data:
test_date
2018-07-01
2018-07-02
...
2019-06-30
2019-07-01
2019-07-02
...
2020-06-30
2020-07-01
I want to increment a row_number value every time right(test_date,5) = '07-01' so that my final result looks like this:
test_date row_num
2018-07-01 1
2018-07-02 1
... 1
2019-06-30 1
2019-07-01 2
2019-07-02 2
... 2
2020-06-30 2
2020-07-01 3
I tried doing something like this:
, ROW_NUMBER() OVER (
PARTITION BY CASE WHEN RIGHT(a.[test_date],5) = '07-01' THEN 1 ELSE 0 END
ORDER BY a.[test_date]
) AS [test2]
But that did not work out for me.
Any suggestions?
Use datepart to identify the correct date, and then add 1 to a sum every time it changes (assuming there will never be more than 1 row per date).
declare #Test table (test_date date);
insert into #Test (test_date)
values
('2018-07-01'),
('2018-07-02'),
('2019-06-30'),
('2019-07-01'),
('2019-07-02'),
('2020-06-30'),
('2020-07-01');
select *
, sum(case when datepart(month,test_date) = 7 and datepart(day,test_date) = 1 then 1 else 0 end) over (order by test_date asc) row_num
from #Test
order by test_date asc;
Returns:
test_date
row_num
2018-07-01
1
2018-07-02
1
2019-06-30
1
2019-07-01
2
2019-07-02
2
2020-06-30
2
2020-07-01
3
You can do it with DENSE_RANK() window function if you subtract 6 months from your dates:
SELECT test_date,
DENSE_RANK() OVER (ORDER BY YEAR(DATEADD(month, -6, test_date))) row_num
FROM tablename
See the demo.
Results:
test_date | row_num
---------- | -------
2018-07-01 | 1
2018-07-02 | 1
2019-06-30 | 1
2019-07-01 | 2
2019-07-02 | 2
2020-06-30 | 2
2020-07-01 | 3
build a running total based on month=7 and day=2
declare #Test table (mykey int,test_date date);
insert into #Test (mykey,test_date)
values
(1,'2018-07-01'),
(2,'2018-07-02'),
(3,'2019-06-30'),
(4,'2019-07-01'),
(5,'2019-07-02'),
(6,'2020-06-30'),
(7,'2020-07-01');
select mykey,test_date,
sum(case when DatePart(Month,test_date)=7 and DatePart(Day,test_date)=2 then 1 else 0 end) over (order by mykey) RunningTotal from #Test
order by mykey
I have a table like below
AID BID CDate
-----------------------------------------------------
1 2 2018-11-01 00:00:00.000
8 1 2018-11-08 00:00:00.000
1 3 2018-11-09 00:00:00.000
7 1 2018-11-15 00:00:00.000
6 1 2018-12-24 00:00:00.000
2 5 2018-11-02 00:00:00.000
2 7 2018-12-15 00:00:00.000
And I am trying to get a result set as follows
ID MaxDate
-------------------
1 2018-12-24 00:00:00.000
2 2018-12-15 00:00:00.000
Each value in the id columns(AID,BID) should return the max of CDate .
ex: in the case of 1, its max CDate is 2018-12-24 00:00:00.000 (here 1 appears under BID)
in the case of 2 , max date is 2018-12-15 00:00:00.000 . (here 2 is under AID)
I tried the following.
1.
select
g.AID,g.BID,
max(g.CDate) as 'LastDate'
from dbo.TT g
inner join
(select AID,BID,max(CDate) as maxdate
from dbo.TT
group by AID,BID)a
on (a.AID=g.AID or a.BID=g.BID)
and a.maxdate=g.CDate
group by g.AID,g.BID
and 2.
SELECT
AID,
CDate
FROM (
SELECT
*,
max_date = MAX(CDate) OVER (PARTITION BY [AID])
FROM dbo.TT
) AS s
WHERE CDate= max_date
Please suggest a 3rd solution.
You can assemble the data in a table expression first, and the compute the max for each value is simple. For example:
select
id, max(cdate)
from (
select aid as id, cdate from t
union all
select bid, cdate from t
) x
group by id
You seem to only care about values that are in both columns. If this interpretation is correct, then:
select id, max(cdate)
from ((select aid as id, cdate, 1 as is_a, 0 as is_b
from t
) union all
(select bid as id, cdate, 1 as is_a, 0 as is_b
from t
)
) ab
group by id
having max(is_a) = 1 and max(is_b) = 1;
Assume this is my table:
ID DATE
--------------
1 2018-11-12
2 2018-11-13
3 2018-11-14
4 2018-11-15
5 2018-11-16
6 2019-03-05
7 2019-05-07
8 2019-05-08
9 2019-05-08
I need to have partitions be determined by the first date in the partition. Where, any date that is within 2 days of the first date, belongs in the same partition.
The table would end up looking like this if each partition was ranked
PARTITION ID DATE
------------------------
1 1 2018-11-12
1 2 2018-11-13
1 3 2018-11-14
2 4 2018-11-15
2 5 2018-11-16
3 6 2019-03-05
4 7 2019-05-07
4 8 2019-05-08
4 9 2019-05-08
I've tried using datediff with lag to compare to the previous date but that would allow a partition to be inappropriately sized based on spacing, for example all of these dates would be included in the same partition:
ID DATE
--------------
1 2018-11-12
2 2018-11-14
3 2018-11-16
4 2018-11-18
3 2018-11-20
4 2018-11-22
Previous flawed attempt:
Mark when a date is more than 2 days past the previous date:
(case when datediff(day, lag(event_time, 1) over (partition by user_id, stage order by event_time), event_time) > 2 then 1 else 0 end)
You need to use a recursive CTE for this, so the operation is expensive.
with t as (
-- add an incrementing column with no gaps
select t.*, row_number() over (order by date) as seqnum
from t
),
cte as (
select id, date, date as mindate, seqnum
from t
where seqnum = 1
union all
select t.id, t.date,
(case when t.date <= dateadd(day, 2, cte.mindate)
then cte.mindate else t.date
end) as mindate,
t.seqnum
from cte join
t
on t.seqnum = cte.seqnum + 1
)
select cte.*, dense_rank() over (partition by mindate) as partition_num
from cte;
I am learning SQL and I was wondering how to select active users by month, depending on their starting and ending date (both timestamp(6)). My table looks like this:
Cust_Num | Start_Date | End_Date
1 | 2018-01-01 | 2019-01-01
2 | 2018-01-01 | NULL
3 | 2019-01-01 | 2019-06-01
4 | 2017-01-01 | 2019-03-01
So, counting the active users by month, I should have an output like:
As of. | Count
2018-06-01 | 3
...
2019-02-01 | 3
2019-07-01 | 1
So far, I do a manual operation by entering each month:
Select
201906,
count(distinct a.cust_num)
From
active_users a
Where
to_date(‘20190630’,’yyyymmdd) between a.start_date and nvl (a.end_date, ‘31-dec-9999)
union all
Select
201905,
count(distinct a.cust_num)
From
active_users a
Where
to_date(‘20190531’,’yyyymmdd) between a.start_date and nvl (a.end_date, ‘31-dec-9999)
union all
...
Not very optimized and sustainable if I want to enter 10 years ao 120 months lol.
Any help is welcome. Thanks a lot!
This query shows the active-user-count effective as-of the end of the month.
How it works:
Convert each input row (with StartDate and EndDate value) into two rows that represent a point-in-time when the active-user-count incremented (on StartDate) and decremented (on EndDate). We need to convert NULL to a far-off date value because NULL values are sorted before instead of after non-NULL values:
This makes your data look like this:
OnThisDate Change
2018-01-01 1
2019-01-01 -1
2018-01-01 1
9999-12-31 -1
2019-01-01 1
2019-06-01 -1
2017-01-01 1
2019-03-01 -1
Then we simply SUM OVER the Change values (after sorting) to get the active-user-count as of that specific date:
So first, sort by OnThisDate:
OnThisDate Change
2017-01-01 1
2018-01-01 1
2018-01-01 1
2019-01-01 1
2019-01-01 -1
2019-03-01 -1
2019-06-01 -1
9999-12-31 -1
Then SUM OVER:
OnThisDate ActiveCount
2017-01-01 1
2018-01-01 2
2018-01-01 3
2019-01-01 4
2019-01-01 3
2019-03-01 2
2019-06-01 1
9999-12-31 0
Then we PARTITION (not group!) the rows by month and sort them by their date so we can identify the last ActiveCount row for that month (this actually happens in the WHERE of the outermost query, using ROW_NUMBER() and COUNT() for each month PARTITION):
OnThisDate ActiveCount IsLastInMonth
2017-01-01 1 1
2018-01-01 2 0
2018-01-01 3 1
2019-01-01 4 0
2019-01-01 3 1
2019-03-01 2 1
2019-06-01 1 1
9999-12-31 0 1
Then filter on that where IsLastInMonth = 1 (actually, where ROW_COUNT() = COUNT(*) inside each PARTITION) to give us the final output data:
At-end-of-month Active-count
2017-01 1
2018-01 3
2019-01 3
2019-03 2
2019-06 1
9999-12 0
This does result in "gaps" in the result-set because the At-end-of-month column only shows rows where the Active-count value actually changed rather than including all possible calendar months - but that's ideal (as far as I'm concerned) because it excludes redundant data. Filling in the gaps can be done inside your application code by simply repeating output rows for each additional month until it reaches the next At-end-of-month value.
Here's the query using T-SQL on SQL Server (I don't have access to Oracle right now). And here's the SQLFiddle I used to come to a solution: http://sqlfiddle.com/#!18/ad68b7/24
SELECT
OtdYear,
OtdMonth,
ActiveCount
FROM
(
-- This query adds columns to indicate which row is the last-row-in-month ( where RowInMonth == RowsInMonth )
SELECT
OnThisDate,
OtdYear,
OtdMonth,
ROW_NUMBER() OVER ( PARTITION BY OtdYear, OtdMonth ORDER BY OnThisDate ) AS RowInMonth,
COUNT(*) OVER ( PARTITION BY OtdYear, OtdMonth ) AS RowsInMonth,
ActiveCount
FROM
(
SELECT
OnThisDate,
YEAR( OnThisDate ) AS OtdYear,
MONTH( OnThisDate ) AS OtdMonth,
SUM( [Change] ) OVER ( ORDER BY OnThisDate ASC ) AS ActiveCount
FROM
(
SELECT
StartDate AS [OnThisDate],
1 AS [Change]
FROM
tbl
UNION ALL
SELECT
ISNULL( EndDate, DATEFROMPARTS( 9999, 12, 31 ) ) AS [OnThisDate],
-1 AS [Change]
FROM
tbl
) AS sq1
) AS sq2
) AS sq3
WHERE
RowInMonth = RowsInMonth
ORDER BY
OtdYear,
OtdMonth
This query can be flattened into fewer nested queries by using aggregate and window functions directly instead of using aliases (like OtdYear, ActiveCount, etc) but that would make the query much harder to understand.
I have created the query which will give the result of all the months starting from the minimum start date in the table till maximum end date.
You can change it using adding one condition in WHERE clause.
-- table creation
CREATE TABLE ACTIVE_USERS (CUST_NUM NUMBER, START_DATE DATE, END_DATE DATE)
-- data creation
INSERT INTO ACTIVE_USERS
SELECT * FROM
(
SELECT 1, DATE '2018-01-01', DATE '2019-01-01' FROM DUAL UNION ALL
SELECT 2, DATE '2018-01-01', NULL FROM DUAL UNION ALL
SELECT 3, DATE '2019-01-01', DATE '2019-06-01' FROM DUAL UNION ALL
SELECT 4, DATE '2017-01-01', DATE '2019-03-01' FROM DUAL
)
-- data in the actual table
SELECT * FROM ACTIVE_USERS ORDER BY CUST_NUM;
CUST_NUM START_DATE END_DATE
---------- ---------- ----------
1 2018-01-01 2019-01-01
2 2018-01-01
3 2019-01-01 2019-06-01
4 2017-01-01 2019-03-01
Query to fetch desired result
WITH CTE ( START_DATE, END_DATE ) AS
(
SELECT
ADD_MONTHS( START_DATE, LEVEL - 1 ),
ADD_MONTHS( START_DATE, LEVEL ) - 1
FROM
(
SELECT
MIN( START_DATE ) AS START_DATE,
MAX( END_DATE ) AS END_DATE
FROM
ACTIVE_USERS
)
CONNECT BY LEVEL <= CEIL( MONTHS_BETWEEN( END_DATE, START_DATE ) ) + 1
)
--
--
SELECT
C.START_DATE,
COUNT(1) AS CNT
FROM
CTE C
JOIN ACTIVE_USERS D ON
(
C.END_DATE BETWEEN
D.START_DATE
AND
CASE
WHEN D.END_DATE IS NOT NULL THEN D.END_DATE
ELSE C.END_DATE
END
)
GROUP BY
C.START_DATE
ORDER BY
C.START_DATE;
-- output --
START_DATE CNT
---------- ----------
2017-01-01 1
2017-02-01 1
2017-03-01 1
2017-04-01 1
2017-05-01 1
2017-06-01 1
2017-07-01 1
2017-08-01 1
2017-09-01 1
2017-10-01 1
2017-11-01 1
START_DATE CNT
---------- ----------
2017-12-01 1
2018-01-01 3
2018-02-01 3
2018-03-01 3
2018-04-01 3
2018-05-01 3
2018-06-01 3
2018-07-01 3
2018-08-01 3
2018-09-01 3
2018-10-01 3
START_DATE CNT
---------- ----------
2018-11-01 3
2018-12-01 3
2019-01-01 3
2019-02-01 3
2019-03-01 2
2019-04-01 2
2019-05-01 2
2019-06-01 1
30 rows selected.
Cheers!!
I have the following data and I want to subtract current row from previous row based on the UserID. I tried the code below is not given me what I want
DECLARE #DATETBLE TABLE (UserID INT, Dates DATE)
INSERT INTO #DATETBLE VALUES
(1,'2018-01-01'), (1,'2018-01-02'), (1,'2018-01-03'),(1,'2018-01-13'),
(2,'2018-01-15'),(2,'2018-01-16'),(2,'2018-01-17'), (5,'2018-02-04'),
(5,'2018-02-05'),(5,'2018-02-06'),(5,'2018-02-11'), (5,'2018-02-17')
;with cte as (
select UserID,Dates, row_number() over (order by UserID) as seqnum
from #DATETBLE t
)
select t.UserID,t.Dates, datediff(day,tprev.Dates,t.Dates)as diff
from cte t left outer join
cte tprev
on t.seqnum = tprev.seqnum + 1;
Current Output
UserID Dates diff
1 2018-01-01 NULL
1 2018-01-02 1
1 2018-01-03 1
1 2018-01-13 10
2 2018-01-15 2
2 2018-01-16 1
2 2018-01-17 1
5 2018-02-04 18
5 2018-02-05 1
5 2018-02-06 1
5 2018-02-11 5
5 2018-02-17 6
My Expected Output
UserID Dates diff
1 2018-01-01 NULL
1 2018-01-02 1
1 2018-01-03 1
1 2018-01-13 10
2 2018-01-15 NULL
2 2018-01-16 1
2 2018-01-17 1
5 2018-02-04 NULL
5 2018-02-05 1
5 2018-02-06 1
5 2018-02-11 5
5 2018-02-17 6
Your tag (sql-server-2008) suggests me to use APPLY :
select t.userid, t.dates, datediff(day, t1.dates, t.dates) as diff
from #DATETBLE t outer apply
( select top (1) t1.*
from #DATETBLE t1
where t1.userid = t.userid and
t1.dates < t.dates
order by t1.dates desc
) t1;
If you have SQL Server version 2012 or higher, you could use LAG() with a partition by UserID:
SELECT UserID
, DATEDIFF(dd,COALESCE(LAG_DATES, Dates), Dates) as diff
FROM
(
SELECT UserID
, Dates
, LAG(Dates) OVER (PARTITION BY UserID ORDER BY Dates) as LAG_DATES
FROM #DATETBLE
) exp
This will give you a 0 value instead of a NULL value for the first date in the sequence though.
Since you tagged the post with SQL Server 2008, however, you may need to use a method that doesn't rely on this windowed function.