I have the following table:
pk_positions ass_pos_id underlying entry_date
1 1 abc 2016-03-14
2 1 xyz 2016-03-17
3 tlt 2016-03-18
4 4 ujf 2016-03-21
5 4 dks 2016-03-23
6 4 dqp 2016-03-26
I need to select one row per ass_pos_id which has the earliest entry_date. Rows which do not have a value for ass_pos_id are not included.
In other words, for each non null ass_pos_id group, select the row which has the earliest entry_date
The following is the desired result:
pk_positions ass_pos_id underlying entry_date
1 1 abc 2016-03-14
4 4 ujf 2016-03-21
You could use the row_number window function:
SELECT pk_positions, ass_pos_id, underlying, entry_date
FROM (SELECT pk_positions, ass_pos_id, underlying, entry_date,
ROW_NUMBER() OVER (PARTITION BY ass_pos_id
ORDER BY entry_date ASC) rn
FROM mytable
WHERE ass_pos_id IS NOT NULL) t
WHERE rn = 1
Related
I'm trying to get the number of days difference in dates between the effdate status 0 that follows the most recent status 1
the code below yields the following results
SELECT * FROM
(SELECT FILEKEY, STATUS, EFFDATE FROM ASTATUSHIST
UNION
SELECT FILEKEY, ASTATUS, ASTATUSEFFDATE FROM USERS ) A
ORDER BY 1, 3 DESC
130 0 2019-10-25 00:00:00.000
130 0 2017-03-01 00:00:00.000
130 0 2017-01-01 00:00:00.000
130 1 2005-02-01 00:00:00.000
130 0 2001-03-03 00:00:00.000
130 0 2000-01-30 00:00:00.000
130 0 2000-01-01 00:00:00.000
this code combines 2 tables to get the complete history for a given user.
Ideally I could produce something that looks like this:
130 4352
or
125 null
where the null is filekey without a status 1 or a filekey with a status 1 but without a following status 0
Thanks
In all supported versions of SQL Server, you can use window functions:
with t as (
<your query here>
)
select t.*,
datediff(day, date, next_date) as days_diff
from (select t.*,
row_number() over (partition by filekey, status order by date desc) as seqnum,
lead(date) over (partition by filekey order by date) as next_date
from t
) t
where seqnum = 1;
Trying to get userid recent aggregate value for session_id.
(session_id 3 has two records, recent agg value is 80.00
session_id 4 has four records, recent agg value is 95.00
session_id 6 has three records, recent agg value is 72.00
Table:session_agg
id session_id userid agg date
-- ---------- ------ ----- -------
1 3 11 60.00 1573561586
4 3 11 80.00 1573561586
6 4 11 35.00 1573561749
7 4 11 50.00 1573561751
8 4 11 70.00 1573561912
10 4 11 95.00 1573561921
11 6 14 40.00 1573561945
12 6 14 67.00 1573561967
13 6 14 72.00 1573561978
select id, session_id, userid, agg, date from session_agg
WHERE date IN (select MAX(date) from session_agg GROUP BY session_id) AND
userid = 11
If you want to stick with your current approach, then you need to correlate the session_id in the subquery which checks for the max date for each session:
SELECT id, session_id, userid, add, date
FROM session_agg sa1
WHERE
date = (SELECT MAX(date) FROM session_agg sa2 WHERE sa2.session_id = sa1.session_id) AND
userid = 11;
But, if your version of SQL supports analytic functions, ROW_NUMBER is an easier way to do this:
WITH cte AS (
SELECT *, ROW_NUMBER() OVER (PARTITION BY session_id ORDER BY date DESC) rn
FROM session_agg
)
SELECT id, session_id, userid, add, date
FROM cte
WHERE rn = 1;
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 have some data that looks like this:
CustID EventID TimeStamp
1 17 1/1/15 13:23
1 17 1/1/15 14:32
1 13 1/1/25 14:54
1 13 1/3/15 1:34
1 17 1/5/15 2:54
1 1 1/5/15 3:00
2 17 2/5/15 9:12
2 17 2/5/15 9:18
2 1 2/5/15 10:02
2 13 2/8/15 7:43
2 13 2/8/15 7:50
2 1 2/8/15 8:00
I'm trying to use the row_number function to get it to look like this:
CustID EventID TimeStamp SeqNum
1 17 1/1/15 13:23 1
1 17 1/1/15 14:32 1
1 13 1/1/25 14:54 2
1 13 1/3/15 1:34 2
1 17 1/5/15 2:54 3
1 1 1/5/15 3:00 4
2 17 2/5/15 9:12 1
2 17 2/5/15 9:18 1
2 1 2/5/15 10:02 2
2 13 2/8/15 7:43 3
2 13 2/8/15 7:50 3
2 1 2/8/15 8:00 4
I tried this:
row_number () over
(partition by custID, EventID
order by custID, TimeStamp asc) SeqNum]
but got this back:
CustID EventID TimeStamp SeqNum
1 17 1/1/15 13:23 1
1 17 1/1/15 14:32 2
1 13 1/1/25 14:54 3
1 13 1/3/15 1:34 4
1 17 1/5/15 2:54 5
1 1 1/5/15 3:00 6
2 17 2/5/15 9:12 1
2 17 2/5/15 9:18 2
2 1 2/5/15 10:02 3
2 13 2/8/15 7:43 4
2 13 2/8/15 7:50 5
2 1 2/8/15 8:00 6
how can I get it to sequence based on the change in the EventID?
This is tricky. You need a multi-step process. You need to identify the groups (a difference of row_number() works for this). Then, assign an increasing constant to each group. And then use dense_rank():
select sd.*, dense_rank() over (partition by custid order by mints) as seqnum
from (select sd.*,
min(timestamp) over (partition by custid, eventid, grp) as mints
from (select sd.*,
(row_number() over (partition by custid order by timestamp) -
row_number() over (partition by custid, eventid order by timestamp)
) as grp
from somedata sd
) sd
) sd;
Another method is to use lag() and a cumulative sum:
select sd.*,
sum(case when prev_eventid is null or prev_eventid <> eventid
then 1 else 0 end) over (partition by custid order by timestamp
) as seqnum
from (select sd.*,
lag(eventid) over (partition by custid order by timestamp) as prev_eventid
from somedata sd
) sd;
EDIT:
The last time I used Amazon Redshift it didn't have row_number(). You can do:
select sd.*, dense_rank() over (partition by custid order by mints) as seqnum
from (select sd.*,
min(timestamp) over (partition by custid, eventid, grp) as mints
from (select sd.*,
(row_number() over (partition by custid order by timestamp rows between unbounded preceding and current row) -
row_number() over (partition by custid, eventid order by timestamp rows between unbounded preceding and current row)
) as grp
from somedata sd
) sd
) sd;
Try this code block:
WITH by_day
AS (SELECT
*,
ts::date AS login_day
FROM table_name)
SELECT
*,
login_day,
FIRST_VALUE(login_day) OVER (PARTITION BY userid ORDER BY login_day , userid rows unbounded preceding) AS first_day
FROM by_day