I have data ( int, date , date types)
SELECT * FROM
(
VALUES
(1700171048,'2020-12-21','2021-01-03'),
(1700171048,'2021-01-05','2021-01-12'),
(1700171048,'2021-01-13','2021-01-17'),
(1700171048,'2021-01-18','2021-01-19'),
(1700171048,'2021-01-22','2021-01-27'),
(1700171048,'2021-01-28','2021-02-17')
(1700171049,'2020-12-21','2021-01-03'),
(1700171049,'2021-01-04','2021-01-05'),
(1700171049,'2021-01-06','2021-01-17'),
(1700171049,'2021-01-18','2021-01-19'),
(1700171049,'2021-01-20','2021-01-27'),
(1700171049,'2021-01-28','2021-02-17')
) AS c (id1, st, endt )
I need output( i.e. if start and end dates are continuous then make it part of group )
id1 st endt
1700171048 '2020-12-21' , '2021-01-03'
1700171048 '2021-01-05' , '2021-01-19'
1700171048 '2021-01-22' , '2021-02-17'
1700171049 '2020-12-21' to '2021-02-17'
I tried this, won't work.
select id, case when min(b.st) = max(b.endt) + 1 then min(b.st) end,
case when min(b.endt) = min(b.st) + 1 then max(b.st) end
from c a join c b
group by id
This is a type of gaps-and-islands problem. Use lag() to identify if there is an overlap. Then a cumulative sum of when there is no overlaps and aggregation:
select id1, min(st), max(endt)
from (select t.*,
sum(case when prev_endt >= st + interval '-1 day' then 0 else 1 end) over (partition by id1 order by st) as grp
from (select t.*,
lag(endt) over (partition by id1 order by st) as prev_endt
from t
) t
) t
group by id1, grp;
Here is a db<>fiddle.
Related
i'm new to SQL and i would need an help.
I have a TAB and I need to find for any item B in the TAB the item A with the closest date. In this case the A with 02.09.2021 04:25:30
Date.
Item
07.09.2021 05:02:05
A
06.09.2021 05:01:02
A
05.09.2021 05:00:02
A
04.09.2021 04:59:01
A
03.09.2021 04:58:03
A
02.09.2021 04:56:55
A
02.09.2021 04:33:56
B
02.09.2021 04:25:30
A
WITH CTE(DATE,ITEM)AS
(
SELECT '20210907 05:02:05' , 'A'UNION ALL
SELECT '20210906 05:01:02' , 'A'UNION ALL
SELECT '20210905 05:00:02' , 'A'UNION ALL
SELECT'20210904 04:59:01' , 'A'UNION ALL
SELECT'20210903 04:58:03' , 'A'UNION ALL
SELECT'20210902 04:56:55' , 'A'UNION ALL
SELECT'20210902 04:33:56' , 'B'UNION ALL
SELECT'20210902 04:25:30' , 'A'
)
SELECT
CAST(C.DATE AS DATETIME)X_DATE,C.ITEM,Q.CLOSEST
FROM CTE AS C
OUTER APPLY
(
SELECT TOP 1 CAST(X.DATE AS DATETIME)CLOSEST
FROM CTE AS X
WHERE X.ITEM='A'AND CAST(X.DATE AS DATETIME)<CAST(C.DATE AS DATETIME)
ORDER BY CAST(X.DATE AS DATETIME) ASC
)Q
WHERE C.ITEM='B'
You can use OUTER APPLY-approach as in the above query.
Please also take a look that datetime-column (DATE)is written in the ISO-compliant form
Your data has only two columns. If you want the only the closest A timestamp, then the fastest way is probably window functions:
select t.*,
(case when prev_a_date is null then next_a_date
when next_a_date is null then prev_a_date
when datediff(second, prev_a_date, date) <= datediff(second, date, next_a_date) then prev_a_date
else next_a_date
end) as a_date
from (select t.*,
max(case when item = 'A' then date end) over (order by date) as prev_a_date,
min(case when item = 'A' then date end) over (order by date desc) as next_a_date
from t
) t
where item = 'B';
This uses seconds to measure the time difference, but you can use a smaller unit if appropriate.
You can also do this using apply if you have more columns from the "A" rows that you want:
select tb.*, ta.*
from t b outer apply
(select top (1) ta.*
from t ta
where item = 'A'
order by abs(datediff(second, a.date, b.date))
) t
where item = 'B';
I have a table TRANS that contains the following records:
TRANS_ID TRANS_DT QTY
1 01-Aug-2020 5
1 01-Aug-2020 1
1 03-Aug-2020 2
2 02-Aug-2020 1
The expected output:
TRANS_ID TRANS_DT BEGBAL TOTAL END_BAL
1 01-Aug-2020 0 6 6
1 02-Aug-2020 6 0 6
1 03-Aug-2020 6 2 8
2 01-Aug-2020 0 0 0
2 02-Aug-2020 0 1 1
2 03-Aug-2020 1 0 1
Each trans_id starts with a beginning balance of 0 (01-Aug-2020). For succeeding days, the beginning balance is the ending balance of the previous day and so on.
I can create PL/SQL block to create the output. Is it possible to get the output in 1 SQL statement?
Thanks.
Try this following script using CTE-
Demo Here
WITH CTE
AS
(
SELECT DISTINCT A.TRANS_ID,B.TRANS_DT
FROM your_table A
CROSS JOIN (SELECT DISTINCT TRANS_DT FROM your_table) B
),
CTE2
AS
(
SELECT C.TRANS_ID,C.TRANS_DT,SUM(D.QTY) QTY
FROM CTE C
LEFT JOIN your_table D
ON C.TRANS_ID = D.TRANS_ID
AND C.TRANS_DT = D.TRANS_DT
GROUP BY C.TRANS_ID,C.TRANS_DT
ORDER BY C.TRANS_ID,C.TRANS_DT
)
SELECT F.TRANS_ID,F.TRANS_DT,
(
SELECT COALESCE (SUM(QTY), 0) FROM CTE2 E
WHERE E.TRANS_ID = F.TRANS_ID AND E.TRANS_DT < F.TRANS_DT
) BEGBAL,
(
SELECT COALESCE (SUM(QTY), 0) FROM CTE2 E
WHERE E.TRANS_ID = F.TRANS_ID AND E.TRANS_DT = F.TRANS_DT
) TOTAL ,
(
SELECT COALESCE (SUM(QTY), 0) FROM CTE2 E
WHERE E.TRANS_ID = F.TRANS_ID AND E.TRANS_DT <= F.TRANS_DT
) END_BAL
FROM CTE2 F
You can as well do like this (I would assume it's a bit faster): Demo
with
dt_between as (
select mindt + level - 1 as trans_dt
from (select min(trans_dt) as mindt, max(trans_dt) as maxdt from t)
connect by level <= maxdt - mindt + 1
),
dt_for_trans_id as (
select *
from dt_between, (select distinct trans_id from t)
),
qty_change as (
select distinct trans_id, trans_dt,
sum(qty) over (partition by trans_id, trans_dt) as total,
sum(qty) over (partition by trans_id order by trans_dt) as end_bal
from t
right outer join dt_for_trans_id using (trans_id, trans_dt)
)
select
trans_id,
to_char(trans_dt, 'DD-Mon-YYYY') as trans_dt,
nvl(lag(end_bal) over (partition by trans_id order by trans_dt), 0) as beg_bal,
nvl(total, 0) as total,
nvl(end_bal, 0) as end_bal
from qty_change q
order by trans_id, trans_dt
dt_between returns all the days between min(trans_dt) and max(trans_dt) in your data.
dt_for_trans_id returns all these days for each trans_id in your data.
qty_change finds difference for each day (which is TOTAL in your example) and cumulative sum over all the days (which is END_BAL in your example).
The main select takes END_BAL from previous day and calls it BEG_BAL, it also does some formatting of final output.
First of all, you need to generate dates, then you need to aggregate your values by TRANS_DT, and then left join your aggregated data to dates. The easiest way to get required sums is to use analitic window functions:
with dates(dt) as ( -- generating dates between min(TRANS_DT) and max(TRANS_DT) from TRANS
select min(trans_dt) from trans
union all
select dt+1 from dates
where dt+1<=(select max(trans_dt) from trans)
)
,trans_agg as ( -- aggregating QTY in TRANS
select TRANS_ID,TRANS_DT,sum(QTY) as QTY
from trans
group by TRANS_ID,TRANS_DT
)
select -- using left join partition by to get data on daily basis for each trans_id:
dt,
trans_id,
nvl(sum(qty) over(partition by trans_id order by dates.dt range between unbounded preceding and 1 preceding),0) as BEGBAL,
nvl(qty,0) as TOTAL,
nvl(sum(qty) over(partition by trans_id order by dates.dt),0) as END_BAL
from dates
left join trans_agg tr
partition by (trans_id)
on tr.trans_dt=dates.dt;
Full example with sample data:
alter session set nls_date_format='dd-mon-yyyy';
with trans(TRANS_ID,TRANS_DT,QTY) as (
select 1,to_date('01-Aug-2020'), 5 from dual union all
select 1,to_date('01-Aug-2020'), 1 from dual union all
select 1,to_date('03-Aug-2020'), 2 from dual union all
select 2,to_date('02-Aug-2020'), 1 from dual
)
,dates(dt) as ( -- generating dates between min(TRANS_DT) and max(TRANS_DT) from TRANS
select min(trans_dt) from trans
union all
select dt+1 from dates
where dt+1<=(select max(trans_dt) from trans)
)
,trans_agg as ( -- aggregating QTY in TRANS
select TRANS_ID,TRANS_DT,sum(QTY) as QTY
from trans
group by TRANS_ID,TRANS_DT
)
select
dt,
trans_id,
nvl(sum(qty) over(partition by trans_id order by dates.dt range between unbounded preceding and 1 preceding),0) as BEGBAL,
nvl(qty,0) as TOTAL,
nvl(sum(qty) over(partition by trans_id order by dates.dt),0) as END_BAL
from dates
left join trans_agg tr
partition by (trans_id)
on tr.trans_dt=dates.dt;
You can use a recursive query to generate the overall date range, cross join it with the list of distinct tran_id, then bring the table with a left join. The last step is aggregation and window functions:
with all_dates (trans_dt, max_dt) as (
select min(trans_dt), max(trans_dt) from trans group by trans_id
union all
select trans_dt + interval '1' day, max_dt from all_dates where trans_dt < max_dt
)
select
i.trans_id,
d.trans_dt,
coalesce(sum(sum(t.qty)) over(partition by i.trans_id order by d.trans_dt), 0) - coalesce(sum(t.qty), 0) begbal,
coalesce(sum(t.qty), 0) total,
coalesce(sum(sum(t.qty)) over(partition by i.trans_id order by d.trans_dt), 0) endbal
from all_dates d
cross join (select distinct trans_id from trans) i
left join trans t on t.trans_id = i.trans_id and t.trans_dt = d.trans_dt
group by i.trans_id, d.trans_dt
order by i.trans_id, d.trans_dt
I have some prices for the month of January.
Date,Price
1,100
2,100
3,115
4,120
5,120
6,100
7,100
8,120
9,120
10,120
Now, the o/p I need is a non-overlapping date range for each price.
price,from,To
100,1,2
115,3,3
120,4,5
100,6,7
120,8,10
I need to do this using SQL only.
For now, if I simply group by and take min and max dates, I get the below, which is an overlapping range:
price,from,to
100,1,7
115,3,3
120,4,10
This is a gaps-and-islands problem. The simplest solution is the difference of row numbers:
select price, min(date), max(date)
from (select t.*,
row_number() over (order by date) as seqnum,
row_number() over (partition by price, order by date) as seqnum2
from t
) t
group by price, (seqnum - seqnum2)
order by min(date);
Why this works is a little hard to explain. But if you look at the results of the subquery, you will see how the adjacent rows are identified by the difference in the two values.
SELECT Lag.price,Lag.[date] AS [From], MIN(Lead.[date]-Lag.[date])+Lag.[date] AS [to]
FROM
(
SELECT [date],[Price]
FROM
(
SELECT [date],[Price],LAG(Price) OVER (ORDER BY DATE,Price) AS LagID FROM #table1 A
)B
WHERE CASE WHEN Price <> ISNULL(LagID,1) THEN 1 ELSE 0 END = 1
)Lag
JOIN
(
SELECT [date],[Price]
FROM
(
SELECT [date],Price,LEAD(Price) OVER (ORDER BY DATE,Price) AS LeadID FROM [#table1] A
)B
WHERE CASE WHEN Price <> ISNULL(LeadID,1) THEN 1 ELSE 0 END = 1
)Lead
ON Lag.[Price] = Lead.[Price]
WHERE Lead.[date]-Lag.[date] >= 0
GROUP BY Lag.[date],Lag.[price]
ORDER BY Lag.[date]
Another method using ROWS UNBOUNDED PRECEDING
SELECT price, MIN([date]) AS [from], [end_date] AS [To]
FROM
(
SELECT *, MIN([abc]) OVER (ORDER BY DATE DESC ROWS UNBOUNDED PRECEDING ) end_date
FROM
(
SELECT *, CASE WHEN price = next_price THEN NULL ELSE DATE END AS abc
FROM
(
SELECT a.* , b.[date] AS next_date, b.price AS next_price
FROM #table1 a
LEFT JOIN #table1 b
ON a.[date] = b.[date]-1
)AA
)BB
)CC
GROUP BY price, end_date
first punch as in time,
second punch as out time
if possible avoid duplicate punch on same time within a minute
I need to get all in time ,outtime in a row with total hours
like below any format.
I tried below query but can't get my expected output
WITH Level1
AS (
SELECT A.emp_reader_id,
DT
,A.EventCatId
,A.Belongs_to
,ROW_NUMBER() OVER ( PARTITION BY A.Belongs_to,A.emp_reader_id ORDER BY DT ) AS RowNum
FROM dbo.trnevents A
)
,
LEVEL2
AS (-- find the last and next event type for each row
SELECT A.emp_reader_id,A.DT , A.EventCatId ,COALESCE(LastVal.EventCatId, 10) AS LastEvent,
COALESCE(NextVal.EventCatId, 10) AS NextEvent ,A.Belongs_to
FROM Level1 A
LEFT JOIN Level1 LastVal
ON A.emp_reader_id = LastVal.emp_reader_id and A.Belongs_to=LastVal.Belongs_to
AND A.RowNum - 1 = LastVal.RowNum
LEFT JOIN Level1 NextVal
ON A.emp_reader_id = NextVal.emp_reader_id and A.Belongs_to=NextVal.Belongs_to
AND A.RowNum + 1 = NextVal.RowNum
)
select * from level2 where emp_reader_id=92 order by dt desc
Expected output:
Try this below script. I considered all DT with Sam Minutes as single entry for the calculation.
WITH CTE AS
(
SELECT MAX(emp_reader_id) emp_reader_id,
CAST(DT AS DATE) Date_for_Group,
LEFT(CAST(DT AS VARCHAR),16) Time_For_Group,
ROW_NUMBER() OVER(PARTITION BY CAST(DT AS DATE) ORDER BY LEFT(CAST(DT AS VARCHAR),16)) RN,
CASE
WHEN ROW_NUMBER() OVER(PARTITION BY CAST(DT AS DATE) ORDER BY LEFT(CAST(DT AS VARCHAR),16))%2 = 0 THEN 'OUT'
ELSE 'IN'
END In_Out
FROM your_table
GROUP BY CAST(DT AS DATE),LEFT(CAST(DT AS VARCHAR),16)
)
SELECT A.emp_reader_id,A.Date_for_Group,
SUM(DATEDIFF(Minute,CAST(A.Time_For_Group AS DATETIME),CAST(B.Time_For_Group AS DATETIME)))/60 Hr,
SUM(DATEDIFF(Minute,CAST(A.Time_For_Group AS DATETIME),CAST(B.Time_For_Group AS DATETIME)))%60 Min
FROM CTE A
INNER JOIN CTE B
ON A.emp_reader_id = B.emp_reader_id
AND A.RN = B.RN -1
AND A.Date_for_Group = B.Date_for_Group
WHERE A.In_Out = 'IN'
GROUP BY A.emp_reader_id,A.Date_for_Group
first assign rownumber to datetime column then start the same result set with rownumber+1
Then Inner join them on rownumbers. After that select min an max from timein and out columns and group by on date to get total workhours of that day. hope it helps.
select empid
,date
,min(timein) as timein,max (timeout) timeout,convert(nvarchar(20),datediff(hh,min (timein),max(timeout))%24)
+':'+
convert(nvarchar(20),datediff(mi,min (timein),max(timeout))%60) as totalhrs
from(
Select a.empid,cast(a.dt as date) date,b.dt as timein,a.dt as timeout from(
SELECT DT
,[empid]
, id
,row_number() over(order by dt) as inn
FROM [test1].[dbo].[Table_2]
)a
inner join(
SELECT distinct DT
,[empid]
, id
,rank() over(order by dt)+1 as out
FROM [test1].[dbo].[Table_2])b
on FORMAT(a.dt,'hh:mm') <> FORMAT(b.dt,'hh:mm')
and cast(a.dt as date)=cast(b.dt as date)
and a.inn=b.out)b
group by b.empid,b.date
I have a table that has data like following.
attr |time
----------------|--------------------------
abc |2018-08-06 10:17:25.282546
def |2018-08-06 10:17:25.325676
pqr |2018-08-05 10:17:25.366823
abc |2018-08-06 10:17:25.407941
def |2018-08-05 10:17:25.449249
I want to group them and count by attr column row wise and also create additional columns in to show their counts per day and percentages as shown below.
attr |day1_count| day1_%| day2_count| day2_%
----------------|----------|-------|-----------|-------
abc |2 |66.6% | 0 | 0.0%
def |1 |33.3% | 1 | 50.0%
pqr |0 |0.0% | 1 | 50.0%
I'm able to display one count by using group by but unable to find out how to even seperate them to multiple columns. I tried to generate day1 percentage with
SELECT attr, count(attr), count(attr) / sum(sub.day1_count) * 100 as percentage from (
SELECT attr, count(*) as day1_count FROM my_table WHERE DATEPART(week, time) = DATEPART(day, GETDate()) GROUP BY attr) as sub
GROUP BY attr;
But this also is not giving me correct answer, I'm getting all zeroes for percentage and count as 1. Any help is appreciated. I'm trying to do this in Redshift which follows postgresql syntax.
Let's nail the logic before presenting:
with CTE1 as
(
select attr, DATEPART(day, time) as theday, count(*) as thecount
from MyTable
)
, CTE2 as
(
select theday, sum(thecount) as daytotal
from CTE1
group by theday
)
select t1.attr, t1.theday, t1.thecount, t1.thecount/t2.daytotal as percentofday
from CTE1 t1
inner join CTE2 t2
on t1.theday = t2.theday
From here you can pivot to create a day by day if you feel the need
I am trying to enhance the query #johnHC btw if you needs for 7days then you have to those days in case when
with CTE1 as
(
select attr, time::date as theday, count(*) as thecount
from t group by attr,time::date
)
, CTE2 as
(
select theday, sum(thecount) as daytotal
from CTE1
group by theday
)
,
CTE3 as
(
select t1.attr, EXTRACT(DOW FROM t1.theday) as day_nmbr,t1.theday, t1.thecount, t1.thecount/t2.daytotal as percentofday
from CTE1 t1
inner join CTE2 t2
on t1.theday = t2.theday
)
select CTE3.attr,
max(case when day_nmbr=0 then CTE3.thecount end) as day1Cnt,
max(case when day_nmbr=0 then percentofday end) as day1,
max(case when day_nmbr=1 then CTE3.thecount end) as day2Cnt,
max( case when day_nmbr=1 then percentofday end) day2
from CTE3 group by CTE3.attr
http://sqlfiddle.com/#!17/54ace/20
In case that you have only 2 days:
http://sqlfiddle.com/#!17/3bdad/3 (days descending as in your example from left to right)
http://sqlfiddle.com/#!17/3bdad/5 (days ascending)
The main idea is already mentioned in the other answers. Instead of joining the CTEs for calculating the values I am using window functions which is a bit shorter and more readable I think. The pivot is done the same way.
SELECT
attr,
COALESCE(max(count) FILTER (WHERE day_number = 0), 0) as day1_count, -- D
COALESCE(max(percent) FILTER (WHERE day_number = 0), 0) as day1_percent,
COALESCE(max(count) FILTER (WHERE day_number = 1), 0) as day2_count,
COALESCE(max(percent) FILTER (WHERE day_number = 1), 0) as day2_percent
/*
Add more days here
*/
FROM(
SELECT *, (count::float/count_per_day)::decimal(5, 2) as percent -- C
FROM (
SELECT DISTINCT
attr,
MAX(time::date) OVER () - time::date as day_number, -- B
count(*) OVER (partition by time::date, attr) as count, -- A
count(*) OVER (partition by time::date) as count_per_day
FROM test_table
)s
)s
GROUP BY attr
ORDER BY attr
A counting the rows per day and counting the rows per day AND attr
B for more readability I convert the date into numbers. Here I take the difference between current date of the row and the maximum date available in the table. So I get a counter from 0 (first day) up to n - 1 (last day)
C calculating the percentage and rounding
D pivot by filter the day numbers. The COALESCE avoids the NULL values and switched them into 0. To add more days you can multiply these columns.
Edit: Made the day counter more flexible for more days; new SQL Fiddle
Basically, I see this as conditional aggregation. But you need to get an enumerator for the date for the pivoting. So:
SELECT attr,
COUNT(*) FILTER (WHERE day_number = 1) as day1_count,
COUNT(*) FILTER (WHERE day_number = 1) / cnt as day1_percent,
COUNT(*) FILTER (WHERE day_number = 2) as day2_count,
COUNT(*) FILTER (WHERE day_number = 2) / cnt as day2_percent
FROM (SELECT attr,
DENSE_RANK() OVER (ORDER BY time::date DESC) as day_number,
1.0 * COUNT(*) OVER (PARTITION BY attr) as cnt
FROM test_table
) s
GROUP BY attr, cnt
ORDER BY attr;
Here is a SQL Fiddle.