I have a dataset that looks like this:
ID HoursWorked TotalHours
23 1 1
23 1 2
23 1 3
23 0.5 3.5
23 1 4.5
23 1 5.5
23 1 6.5
23 1 7.5
23 1 8.5
61 1 1
61 1 2
What I want to do is if the total hours hits 8 hours, I want to split that row (e.g. 8.5 in the sample data above) so that an employee always has the total hours of 8. If someone works over 8 hours it should continue after hitting 8 in the totalhours column. For example, I want something like this as my final result.
ID HoursWorked TotalHours
23 1 1
23 1 2
23 1 3
23 0.5 3.5
23 1 4.5
23 1 5.5
23 1 6.5
23 1 7.5
23 0.5 8 *
23 0.5 8.5 *
61 1 1
61 1 2
As you can see the row which originally had 8.5 for its totalhours got broken down into two different rows.
I couldn't think of any way to do this in SQL Server. I'd appreciate any help on this.
see if this works.
select ID,HoursWorked,TotalHours from table_name where TotalHours <=8
union
select ID,(HoursWorked-(TotalHours-8) as HoursWorked ,8 as TotalHours from table_name where TotalHours >8
union
select ID,(TotalHours-8) as HoursWorked ,TotalHours from table_name where TotalHours >8
This seems rather complicated. This approach takes all the rows before 8 hours. It then finds the row that first passes 8 hours and splits that one as needed:
select id, hoursworked, totalhours
from t
where totalhours <= 8
union all
select t.id, v.hoursworked, v.totalhours
from (select t.*, row_number() over (partition by id order by totalhours) as seqnum
from t
where totalhours > 8
) t cross apply
(values (case when seqnum = 1 then totalhours - 8 end,
case when seqnum = 1 then 8 end
),
(case when seqnum = 1 and totalhours >= 8 then totalhours - 8 else hoursworked end,
totalhours
)
) v(hoursworked, totalhours)
where v.hoursworked > 0
order by id, totalhours;
Here is a db<>fiddle.
Related
I want to group orders from the same customer if they happen within 10 minutes of the first order, then find the next first order and group them and so on.
Ex:
Customer group orders
6 1 3
2 4,5
3 8
7 1 9,10
2 11,12
3 13
id customer time
3 6 2021-05-12 12:14:22.000000
4 6 2021-05-12 12:24:24.000000
5 6 2021-05-12 12:29:16.000000
8 6 2021-05-12 13:01:40.000000
9 7 2021-05-14 12:13:11.000000
10 7 2021-05-14 12:20:01.000000
11 7 2021-05-14 12:45:00.000000
12 7 2021-05-14 12:48:41.000000
13 7 2021-05-14 12:58:16.000000
18 9 2021-05-18 12:22:13.000000
25 15 2021-05-18 13:44:02.000000
26 16 2021-05-17 09:39:02.000000
27 16 2021-05-18 19:38:43.000000
28 17 2021-05-18 15:40:02.000000
29 18 2021-05-19 15:32:53.000000
30 18 2021-05-19 15:45:56.000000
31 18 2021-05-19 16:29:09.000000
34 15 2021-05-24 15:45:14.000000
35 15 2021-05-24 15:45:14.000000
36 19 2021-05-24 17:14:53.000000
Here is what I have currently, I think that it is currently not grouping by customer when case when d.StartTime > dateadd(minute, 10, c.first_time) so it compares StartTime of all orders for all customers.
with
data as (select Customer,StartTime,Id, row_number() over(partition by Customer order by StartTime) rn from orders t),
cte as (
select d.*, StartTime as first_time
from data d
where rn = 1
union all
select d.*,
case when d.StartTime > dateadd(minute, 10, c.first_time)
then d.StartTime
else c.first_time
end
from cte c
inner join data d on d.rn = c.rn + 1
)
select c.*, dense_rank() over(partition by Customer order by first_time) grp
from cte c;'
I have two databases (MySQL & SQL Server) having similar schema so either would work for me.
Try the following on SQL Server:
SELECT customer,
ROW_NUMBER() OVER (PARTITION BY customer ORDER BY grp) AS group_no,
STRING_AGG(id, ',') AS orders
FROM
(
SELECT id,customer, [time],
(DATEDIFF(SECOND, MIN([time]) OVER (PARTITION BY CUSTOMER), [time])/60)/10 grp
FROM orders
) T
GROUP BY customer, grp
ORDER BY customer
See a demo.
According to your posted requirement, you are trying to divide the period between the first order date and the last order date into groups (or let's say time frames) each one is 10 minutes long.
What I did in this query: for each customer order, find the difference between the order date and the minimum date (first customer order date) in seconds and then divide it by 10 to get it's time frame number. i.e. for a difference = 599s the frame number = 599/60 =9m /10 = 0. for a difference = 620s the frame number = 620/60 =10m /10 = 1.
After defining the correct groups/time frames for each order you can simply use the STRING_AGG function to get the desired output. Noting that the STRING_AGG function applies to SQL Server 2017 (14.x) and later.
I want to count the distinct amount of users over the last 60 days, and then, count the distinct amount of users over the last 59 days, and so on and so forth.
Ideally, the output would look like this (TARGET OUTPUT)
Day Distinct Users
60 200
59 200
58 188
57 185
56 180
[...] [...]
where 60 days is the max total possible distinct users, and then 59 would have a little less and so on and so forth.
my query looks like this.
select
count(distinct (case when datediff(day,DATE,current_date) <= 60 then USER_ID end)) as day_60,
count(distinct (case when datediff(day,DATE,current_date) <= 59 then USER_ID end)) as day_59,
count(distinct (case when datediff(day,DATE,current_date) <= 58 then USER_ID end)) as day_58
FROM Table
The issue with my query is that This outputs the data by column instead of by rows (like shown below) AND, most importantly, I have to write out this logic 60x for each of the 60 days.
Current Output:
Day_60 Day_59 Day_58
209 207 207
Is it possible to write the SQL in a way that creates the target as shown initially above?
Using below data in CTE format -
with data_cte(dates,userid) as
(select * from values
('2022-05-01'::date,'UID1'),
('2022-05-01'::date,'UID2'),
('2022-05-02'::date,'UID1'),
('2022-05-02'::date,'UID2'),
('2022-05-03'::date,'UID1'),
('2022-05-03'::date,'UID2'),
('2022-05-03'::date,'UID3'),
('2022-05-04'::date,'UID1'),
('2022-05-04'::date,'UID1'),
('2022-05-04'::date,'UID2'),
('2022-05-04'::date,'UID3'),
('2022-05-04'::date,'UID4'),
('2022-05-05'::date,'UID1'),
('2022-05-06'::date,'UID1'),
('2022-05-07'::date,'UID1'),
('2022-05-07'::date,'UID2'),
('2022-05-08'::date,'UID1')
)
Query to get all dates and count and distinct counts -
select dates,count(userid) cnt, count(distinct userid) cnt_d
from data_cte
group by dates;
DATES
CNT
CNT_D
2022-05-01
2
2
2022-05-02
2
2
2022-05-03
3
3
2022-05-04
5
4
2022-05-05
1
1
2022-05-06
1
1
2022-05-08
1
1
2022-05-07
2
2
Query to get difference of date from current date
select dates,datediff(day,dates,current_date()) ddiff,
count(userid) cnt,
count(distinct userid) cnt_d
from data_cte
group by dates;
DATES
DDIFF
CNT
CNT_D
2022-05-01
45
2
2
2022-05-02
44
2
2
2022-05-03
43
3
3
2022-05-04
42
5
4
2022-05-05
41
1
1
2022-05-06
40
1
1
2022-05-08
38
1
1
2022-05-07
39
2
2
Get records with date difference beyond a certain range only -
include clause having
select datediff(day,dates,current_date()) ddiff,
count(userid) cnt,
count(distinct userid) cnt_d
from data_cte
group by dates
having ddiff<=43;
DDIFF
CNT
CNT_D
43
3
3
42
5
4
41
1
1
39
2
2
38
1
1
40
1
1
If you need to prefix 'day' to each date diff count, you can
add and outer query to previously fetched data-set and add the needed prefix to the date diff column as following -
I am using CTE syntax, but you may use sub-query given you will select from table -
,cte_1 as (
select datediff(day,dates,current_date()) ddiff,
count(userid) cnt,
count(distinct userid) cnt_d
from data_cte
group by dates
having ddiff<=43)
select 'day_'||to_char(ddiff) days,
cnt,
cnt_d
from cte_1;
DAYS
CNT
CNT_D
day_43
3
3
day_42
5
4
day_41
1
1
day_39
2
2
day_38
1
1
day_40
1
1
Updated the answer to get distinct user count for number of days range.
A clause can be included in the final query to limit to number of days needed.
with data_cte(dates,userid) as
(select * from values
('2022-05-01'::date,'UID1'),
('2022-05-01'::date,'UID2'),
('2022-05-02'::date,'UID1'),
('2022-05-02'::date,'UID2'),
('2022-05-03'::date,'UID5'),
('2022-05-03'::date,'UID2'),
('2022-05-03'::date,'UID3'),
('2022-05-04'::date,'UID1'),
('2022-05-04'::date,'UID6'),
('2022-05-04'::date,'UID2'),
('2022-05-04'::date,'UID3'),
('2022-05-04'::date,'UID4'),
('2022-05-05'::date,'UID7'),
('2022-05-06'::date,'UID1'),
('2022-05-07'::date,'UID8'),
('2022-05-07'::date,'UID2'),
('2022-05-08'::date,'UID9')
),cte_1 as
(select datediff(day,dates,current_date()) ddiff,userid
from data_cte), cte_2 as
(select distinct ddiff from cte_1 )
select cte_2.ddiff,
(select count(distinct userid)
from cte_1 where cte_1.ddiff <= cte_2.ddiff) cnt
from cte_2
order by cte_2.ddiff desc
DDIFF
CNT
47
9
46
9
45
9
44
8
43
5
42
4
41
3
40
1
You can do unpivot after getting your current output.
sample one.
select
*
from (
select
209 Day_60,
207 Day_59,
207 Day_58
)unpivot ( cnt for days in (Day_60,Day_59,Day_58));
My database is an amazon redshift.
I have a table that looks like this -
id
nested_id
date
value
1
10
'2021-01-01'
5
1
20
'2021-01-01'
10
1
10
'2021-01-02'
6
1
20
'2021-01-02'
11
1
10
'2021-01-03'
7
1
20
'2021-01-03'
12
2
30
'2021-01-01'
5
2
40
'2021-01-01'
10
2
30
'2021-01-02'
6
2
40
'2021-01-02'
11
2
30
'2021-01-03'
7
2
40
'2021-01-03'
12
So this is basically a table that tracks values by id over time, except for every id there can be a nested_id. And the dates and values are primarily connected to the nested_id.
However, let's say I'm starting with the id field, but for each id I want to only return the points over time for the nested_id that has the greater sum of points.
So right now I'm just grabbing it like this...
select *
from mytable
where id in (1, 2)
except I only want it to return nested_id rows where the maximum value of that nested_id is the greatest.
So here's how I would do this manually.
For id of 1, the maximum value is 12, and the nested_id of that value is 20
For id of 2, the maximum value is 12, and the nested_id of that value is 40
So my return table should be
id
nested_id
date
value
1
20
'2021-01-01'
10
1
20
'2021-01-02'
11
1
20
'2021-01-03'
12
2
40
'2021-01-01'
10
2
40
'2021-01-02'
11
2
40
'2021-01-03'
12
Is there an easy way of performing this query? I'm assuming you have to partition somehow?
You can solve this with row_number window functions
with maxs as (
select id,
nested_id,
value,
row_number() over (partition by id order by value desc) rn
from mytable
)
select mt.*
from mytable mt
left join maxs on mt.id = maxs.id and mt.nested_id = maxs.nested_id
where maxs.rn = 1
I have a table like this:
ID Date Prod
1 1/1/2009 5
1 2/1/2009 5
1 3/1/2009 5
1 4/1/2009 5
1 5/1/2009 5
1 6/1/2009 5
1 7/1/2009 5
1 8/1/2009 5
1 9/1/2009 5
And I need to get the following result:
ID Date Prod CumProd
1 2009/03/01 5 15 ---Each 3 months
1 2009/06/01 5 30 ---Each 3 months
1 2009/09/01 5 45 ---Each 3 months
What could be the best approach to take in SQL?
You can try the below - using window function
DEMO Here
select * from
(
select *,sum(prod) over(order by DATEPART(qq,dateval)) as cum_sum,
row_number() over(partition by DATEPART(qq,dateval) order by dateval) as rn
from t
)A where rn=1
How about just filtering on the month number?
select t.*
from (select id, date, prod, sum(prod) over (partition by id order by date) as running_prod
from t
) t
where month(date) in (3, 6, 9, 12);
I'm stuck with a seemingly easy query, but couldn't manage to get it working the last hours.
I have a table files that holds file names and some values like records in this file, DATE of creation (create_date), DATE of processing (processing_date) and so on. There can be multiple files for a create date in different hours and it is likely that they will not get processed in the same day of creaton, in fact it can even take up to three days or longer for them to get processed.
So let's assume I have these rows, as an example:
create_date | processing_date
------------------------------
2012-09-10 11:10:55.0 | 2012-09-11 18:00:18.0
2012-09-10 15:20:18.0 | 2012-09-11 13:38:19.0
2012-09-10 19:30:48.0 | 2012-09-12 10:59:00.0
2012-09-11 08:19:11.0 | 2012-09-11 18:14:44.0
2012-09-11 22:31:42.0 | 2012-09-21 03:51:09.0
What I want in a single query is to get a grouped column truncated to the day create_date with 11 additional columns for the differences between the processing_date and the create_date, so that the result should roughly look like this:
create_date | diff0days | diff1days | diff2days | ... | diff10days
------------------------------------------------------------------------
2012-09-10 | 0 2 1 ... 0
2012-09-11 | 1 0 0 ... 1
and so on, I hope you get the point :)
I have tried this and so far it works getting a single aggregated column for a create_date with a difference of - for example - 3:
SELECT TRUNC(f.create_date, 'DD') as created, count(1) FROM files f WHERE TRUNC(f.process_date, 'DD') - trunc(f.create_date, 'DD') = 3 GROUP BY TRUNC(f.create_date, 'DD')
I tried combining the single queries and I tried sub-queries, but that didn't help or at least my knowledge about SQL is not sufficient.
What I need is a hint so that I can include the various differences as columns, like shown above. How could I possibly achieve this?
That's basically the pivoting problem:
SELECT TRUNC(f.create_date, 'DD') as created
, sum(case TRUNC(f.process_date, 'DD') - trunc(f.create_date, 'DD')
when 0 then 1 end) as diff0days
, sum(case TRUNC(f.process_date, 'DD') - trunc(f.create_date, 'DD')
when 1 then 1 end) as diff1days
, sum(case TRUNC(f.process_date, 'DD') - trunc(f.create_date, 'DD')
when 2 then 1 end) as diff2days
, ...
FROM files f
GROUP BY
TRUNC(f.create_date, 'DD')
SELECT CreateDate,
sum(CASE WHEN DateDiff(day, CreateDate, ProcessDate) = 1 THEN 1 ELSE 0 END) AS Diff1,
sum(CASE WHEN DateDiff(day, CreateDate, ProcessDate) = 2 THEN 1 ELSE 0 END) AS Diff2,
...
FROM table
GROUP BY CreateDate
ORDER BY CreateDate
As you are using Oracle 11g you can also get desired result by using pivot query.
Here is an example:
-- sample of data from your question
SQL> create table Your_table(create_date, processing_date) as
2 (
3 select '2012-09-10', '2012-09-11' from dual union all
4 select '2012-09-10', '2012-09-11' from dual union all
5 select '2012-09-10', '2012-09-12' from dual union all
6 select '2012-09-11', '2012-09-11' from dual union all
7 select '2012-09-11', '2012-09-21' from dual
8 )
9 ;
Table created
SQL> with t2 as(
2 select create_date
3 , processing_date
4 , to_date(processing_date, 'YYYY-MM-DD')
- To_Date(create_date, 'YYYY-MM-DD') dif
5 from your_table
6 )
7 select create_date
8 , max(diff0) diff0
9 , max(diff1) diff1
10 , max(diff2) diff2
11 , max(diff3) diff3
12 , max(diff4) diff4
13 , max(diff5) diff5
14 , max(diff6) diff6
15 , max(diff7) diff7
16 , max(diff8) diff8
17 , max(diff9) diff9
18 , max(diff10) diff10
19 from (select *
20 from t2
21 pivot(
22 count(dif)
23 for dif in ( 0 diff0
24 , 1 diff1
25 , 2 diff2
26 , 3 diff3
27 , 4 diff4
28 , 5 diff5
29 , 6 diff6
30 , 7 diff7
31 , 8 diff8
32 , 9 diff9
33 , 10 diff10
34 )
35 ) pd
36 ) res
37 group by create_date
38 ;
Result:
Create_Date Diff0 Diff1 Diff2 Diff3 Diff4 Diff5 Diff6 Diff7 Diff8 Diff9 Diff10
--------------------------------------------------------------------------------
2012-09-10 0 2 1 0 0 0 0 0 0 0 0
2012-09-11 1 0 0 0 0 0 0 0 0 0 1