I am using Oracle and trying to retrieve the total number of days a person was out of the office during the year. I have 2 tables involved:
Statuses
1 - Active
2 - Out of the Office
3 - Other
ScheduleHistory
RecordID - primary key
PersonID
PreviousStatusID
NextStatusID
DateChanged
I can easily find when the person went on vacation and when they came back, using
SELECT DateChanged FROM ScheduleHistory WHERE PersonID=111 AND NextStatusID = 2
and
SELECT DateChanged FROM ScheduleHistory WHERE PersonID=111 AND PreviousStatusID = 2
But in case a person went on vacation more than once, how can I can I calculate total number of days a person was out of the office. Is it possible to do programmatically, given only PersonID?
Here is some sample data:
RecordID PersonID PreviousStatusID NextStatusID DateChanged
-----------------------------------------------------------------------------
1 111 1 2 03/11/2020
2 111 2 1 03/13/2020
3 111 1 3 04/01/2020
4 111 3 1 04/07/2020
5 111 1 2 06/03/2020
6 111 2 1 06/05/2020
7 111 1 2 09/14/2020
8 111 2 1 09/17/2020
So from the data above, for the year 2020 for PersonID 111 the query should return 7
Try this:
with aux1 AS (
SELECT
a.*,
to_date(datechanged, 'MM/DD/YYYY') - LAG(to_date(datechanged, 'MM/DD/YYYY')) OVER(
PARTITION BY personid
ORDER BY
recordid
) lag_date
FROM
ScheduleHistory a
)
SELECT
personid,
SUM(lag_date) tot_days_ooo
FROM
aux1
WHERE
previousstatusid = 2
GROUP BY
personid;
If you want total days (or weekdays) for each year (and to account for periods when it goes over the year boundary) then:
WITH date_ranges ( personid, status, start_date, end_date ) AS (
SELECT personid,
nextstatusid,
datechanged,
LEAD(datechanged, 1, datechanged) OVER(
PARTITION BY personid
ORDER BY datechanged
)
FROM table_name
),
split_year_ranges ( personid, year, start_date, end_date, max_date ) AS (
SELECT personid,
TRUNC( start_date, 'YY' ),
start_date,
LEAST(
end_date,
ADD_MONTHS( TRUNC( start_date, 'YY' ), 12 )
),
end_date
FROM date_ranges
WHERE status = 2
UNION ALL
SELECT personid,
end_date,
end_date,
LEAST( max_date, ADD_MONTHS( end_date, 12 ) ),
max_date
FROM split_year_ranges
WHERE end_date < max_date
)
SELECT personid,
EXTRACT( YEAR FROM year) AS year,
SUM( end_date - start_date ) AS total_days,
SUM(
( TRUNC( end_date, 'IW' ) - TRUNC( start_date, 'IW' ) ) * 5 / 7
+ LEAST( end_date - TRUNC( end_date, 'IW' ), 5 )
- LEAST( start_date - TRUNC( start_date, 'IW' ), 5 )
) AS total_weekdays
FROM split_year_ranges
GROUP BY personid, year
ORDER BY personid, year
Which, for the sample data:
CREATE TABLE table_name ( RecordID, PersonID, PreviousStatusID, NextStatusID, DateChanged ) AS
SELECT 1, 111, 1, 2, DATE '2020-03-11' FROM DUAL UNION ALL
SELECT 2, 111, 2, 1, DATE '2020-03-13' FROM DUAL UNION ALL
SELECT 3, 111, 1, 3, DATE '2020-04-01' FROM DUAL UNION ALL
SELECT 4, 111, 3, 1, DATE '2020-04-07' FROM DUAL UNION ALL
SELECT 5, 111, 1, 2, DATE '2020-06-03' FROM DUAL UNION ALL
SELECT 6, 111, 2, 1, DATE '2020-06-05' FROM DUAL UNION ALL
SELECT 7, 111, 1, 2, DATE '2020-09-14' FROM DUAL UNION ALL
SELECT 8, 111, 2, 1, DATE '2020-09-17' FROM DUAL UNION ALL
SELECT 9, 222, 1, 2, DATE '2019-12-31' FROM DUAL UNION ALL
SELECT 10, 222, 2, 2, DATE '2020-12-01' FROM DUAL UNION ALL
SELECT 11, 222, 2, 2, DATE '2021-01-02' FROM DUAL;
Outputs:
PERSONID
YEAR
TOTAL_DAYS
TOTAL_WEEKDAYS
111
2020
7
7
222
2019
1
1
222
2020
366
262
222
2021
1
1
db<>fiddle here
Provided no vacation crosses a year boundary
with grps as (
SELECT sh.*,
row_number() over (partition by PersonID, NextStatusID order by DateChanged) grp
FROM ScheduleHistory sh
WHERE NextStatusID in (1,2) and 3 not in (NextStatusID, PreviousStatusID)
), durations as (
SELECT PersonID, min(DateChanged) DateChanged, max(DateChanged) - min(DateChanged) duration
FROM grps
GROUP BY PersonID, grp
)
SELECT PersonID, sum(duration) days_out
FROM durations
GROUP BY PersonID;
db<>fiddle
year_span is used to split an interval spanning across two years in two different records
H1 adds a row number dependent from PersonID to get the right sequence for each person
H2 gets the periods for each status change and extract 1st day of the year of the interval end
H3 split records that span across two years and calculate the right date_start and date_end for each interval
H calculates days elapsed in each interval for each year
final query sum up the records to get output
EDIT
If you need workdays instead of total days, you should not use total_days/7*5 because it is a bad approximation and in some cases gives weird results.
I have posted a solution to jump on fridays to mondays here
with
statuses (sid, sdescr) as (
select 1, 'Active' from dual union all
select 2, 'Out of the Office' from dual union all
select 3, 'Other' from dual
),
ScheduleHistory(RecordID, PersonID, PreviousStatusID, NextStatusID , DateChanged) as (
select 1, 111, 1, 2, date '2020-03-11' from dual union all
select 2, 111, 2, 1, date '2020-03-13' from dual union all
select 3, 111, 1, 3, date '2020-04-01' from dual union all
select 4, 111, 3, 1, date '2020-04-07' from dual union all
select 5, 111, 1, 2, date '2020-06-03' from dual union all
select 6, 111, 2, 1, date '2020-06-05' from dual union all
select 7, 111, 1, 2, date '2020-09-14' from dual union all
select 8, 111, 2, 1, date '2020-09-17' from dual union all
SELECT 9, 222, 1, 2, date '2019-12-31' from dual UNION ALL
SELECT 10, 222, 2, 2, date '2020-12-01' from dual UNION ALL
SELECT 11, 222, 2, 2, date '2021-01-02' from dual
),
year_span (n) as (
select 1 from dual union all
select 2 from dual
),
H1 AS (
SELECT ROW_NUMBER() OVER (PARTITION BY PersonID ORDER BY RecordID) PID, H.*
FROM ScheduleHistory H
),
H2 as (
SELECT
H1.*, H2.DateChanged DateChanged2,
EXTRACT(YEAR FROM H2.DateChanged) - EXTRACT(YEAR FROM H1.DateChanged) + 1 Y,
trunc(H2.DateChanged,'YEAR') Y2
FROM H1 H1
LEFT JOIN H1 H2 ON H1.PID = H2.PID-1 AND H1.PersonID = H2.PersonID
),
H3 AS (
SELECT Y, N, H2.PID, H2.RecordID, H2.PersonID, H2.NextStatusID,
CASE WHEN Y=1 THEN H2.DateChanged ELSE CASE WHEN N=1 THEN H2.DateChanged ELSE Y2 END END D1,
CASE WHEN Y=1 THEN H2.DateChanged2 ELSE CASE WHEN N=1 THEN Y2 ELSE H2.DateChanged2 END END D2
FROM H2
JOIN year_span N ON N.N <=Y
),
H AS (
SELECT PersonID, NextStatusID, EXTRACT(year FROM d1) Y, d2-d1 D
FROM H3
)
select PersonID, sdescr Status, Y, sum(d) d
from H
join statuses s on NextStatusID = s.sid
group by PersonID, sdescr, Y
order by PersonID, sdescr, Y
output
PersonID Status Y d
111 Active 2020 177
111 Other 2020 6
111 Out of the Office 2020 7
222 Out of the Office 2019 1
222 Out of the Office 2020 366
222 Out of the Office 2021 1
check the fiddle here
Related
I am working in oracle sql. I have two table which is linked to each other by one column - company_id (see on the picture); I want to merge table 1 to table 2 and calculate 6 month average (6 month before period from table 2) of income for each company_id and each date of table2. I appreciate any code/idea how to solve this task.
You can use an analytic range window to calculate the averages for table1 and then JOIN the result to table2:
SELECT t2.*,
t1.avg_income_6,
t1.avg_income_12
FROM table2 t2
LEFT OUTER JOIN (
SELECT company_id,
dt,
ROUND(AVG(income) OVER (
PARTITION BY company_id
ORDER BY dt
RANGE BETWEEN INTERVAL '5' MONTH PRECEDING
AND INTERVAL '0' MONTH FOLLOWING
), 2) AS avg_income_6,
ROUND(AVG(income) OVER (
PARTITION BY company_id
ORDER BY dt
RANGE BETWEEN INTERVAL '11' MONTH PRECEDING
AND INTERVAL '0' MONTH FOLLOWING
), 2) AS avg_income_12
FROM table1
) t1
ON (t2.company_id = t1.company_id AND t2.dt = t1.dt);
Which, for the sample data:
CREATE TABLE table1 (company_id, dt, income) AS
SELECT 1, date '2019-01-01', 65 FROM DUAL UNION ALL
SELECT 1, date '2019-02-01', 58 FROM DUAL UNION ALL
SELECT 1, date '2019-03-01', 12 FROM DUAL UNION ALL
SELECT 1, date '2019-04-01', 81 FROM DUAL UNION ALL
SELECT 1, date '2019-05-01', 38 FROM DUAL UNION ALL
SELECT 1, date '2019-06-01', 81 FROM DUAL UNION ALL
SELECT 1, date '2019-07-01', 38 FROM DUAL UNION ALL
SELECT 1, date '2019-08-01', 69 FROM DUAL UNION ALL
SELECT 1, date '2019-09-01', 54 FROM DUAL UNION ALL
SELECT 1, date '2019-10-01', 90 FROM DUAL UNION ALL
SELECT 1, date '2019-11-01', 10 FROM DUAL UNION ALL
SELECT 1, date '2019-12-01', 12 FROM DUAL UNION ALL
SELECT 1, date '2020-01-01', 11 FROM DUAL UNION ALL
SELECT 1, date '2020-02-01', 83 FROM DUAL UNION ALL
SELECT 1, date '2020-03-01', 18 FROM DUAL UNION ALL
SELECT 1, date '2020-04-01', 28 FROM DUAL UNION ALL
SELECT 1, date '2020-05-01', 52 FROM DUAL UNION ALL
SELECT 1, date '2020-06-01', 21 FROM DUAL UNION ALL
SELECT 1, date '2020-07-01', 54 FROM DUAL UNION ALL
SELECT 1, date '2020-08-01', 30 FROM DUAL UNION ALL
SELECT 1, date '2020-09-01', 12 FROM DUAL UNION ALL
SELECT 1, date '2020-10-01', 25 FROM DUAL UNION ALL
SELECT 1, date '2020-11-01', 86 FROM DUAL UNION ALL
SELECT 1, date '2020-12-01', 4 FROM DUAL UNION ALL
SELECT 1, date '2021-01-01', 10 FROM DUAL UNION ALL
SELECT 1, date '2021-02-01', 72 FROM DUAL UNION ALL
SELECT 1, date '2021-03-01', 65 FROM DUAL UNION ALL
SELECT 1, date '2021-04-01', 25 FROM DUAL;
CREATE TABLE table2 (company_id, dt) AS
SELECT 1, date '2019-06-01' FROM DUAL UNION ALL
SELECT 1, date '2019-09-01' FROM DUAL UNION ALL
SELECT 1, date '2019-12-01' FROM DUAL UNION ALL
SELECT 1, date '2020-01-01' FROM DUAL UNION ALL
SELECT 1, date '2020-07-01' FROM DUAL UNION ALL
SELECT 1, date '2020-08-01' FROM DUAL UNION ALL
SELECT 1, date '2021-03-01' FROM DUAL UNION ALL
SELECT 1, date '2021-04-01' FROM DUAL;
Outputs:
COMPANY_ID
DT
AVG_INCOME_6
AVG_INCOME_12
1
2019-06-01 00:00:00
55.83
55.83
1
2019-09-01 00:00:00
60.17
55.11
1
2019-12-01 00:00:00
45.5
50.67
1
2020-01-01 00:00:00
41
46.17
1
2020-07-01 00:00:00
42.67
41.83
1
2020-08-01 00:00:00
33.83
38.58
1
2021-03-01 00:00:00
43.67
38.25
1
2021-04-01 00:00:00
43.67
38
db<>fiddle here
I don't think you need any window function here (if you were thinking of analytic functions); ordinary avg with appropriate join conditions should do the job.
Sample data:
SQL> with
2 table1 (company_id, datum, income) as
3 (select 1, date '2019-01-01', 65 from dual union all
4 select 1, date '2019-02-01', 58 from dual union all
5 select 1, date '2019-03-01', 12 from dual union all
6 select 1, date '2019-04-01', 81 from dual union all
7 select 1, date '2019-05-01', 38 from dual union all
8 select 1, date '2019-06-01', 81 from dual union all
9 select 1, date '2019-07-01', 38 from dual union all
10 select 1, date '2019-08-01', 69 from dual union all
11 select 1, date '2019-09-01', 54 from dual union all
12 select 1, date '2019-10-01', 90 from dual union all
13 select 1, date '2019-11-01', 10 from dual union all
14 select 1, date '2019-12-01', 12 from dual
15 ),
16 table2 (company_id, datum) as
17 (select 1, date '2019-06-01' from dual union all
18 select 1, date '2019-09-01' from dual union all
19 select 1, date '2019-12-01' from dual union all
20 select 1, date '2020-01-01' from dual union all
21 select 1, date '2020-07-01' from dual
22 )
Query begins here:
23 select b.company_id,
24 b.datum ,
25 round(avg(a.income), 2) result
26 from table1 a join table2 b on a.company_id = b.company_id
27 and a.datum > add_months(b.datum, -6)
28 and a.datum <= b.datum
29 group by b.company_id, b.datum;
COMPANY_ID DATUM RESULT
---------- -------- ----------
1 01.06.19 55,83
1 01.09.19 60,17
1 01.12.19 45,5
1 01.01.20 47
SQL>
I have the below table and I would like to count, day by day, the number of distinct people who logged in everyday. For example, for day 1, everyone logged in, so it's 4. For day 4, there's just one person ID who logged in everyday since day 1, so the count would be 1.
DAY
PERSON_ID
1
01
1
02
1
03
1
04
2
01
2
02
2
03
3
01
4
02
4
01
Expected output.
DAY
PEOPLE_LOGGED_EVERYDAY
PEOPLE
1
4
01, 02, 03, 04
2
3
01, 02, 03
3
1
01
4
1
01
EDIT: the query should also work on the below data.
with t ( DAY, PERSON_ID ) AS(
SELECT 10, '01' FROM DUAL UNION ALL
SELECT 10, '02' FROM DUAL UNION ALL
SELECT 10, '04' FROM DUAL UNION ALL
SELECT 10, '04' FROM DUAL UNION ALL
SELECT 12, '01' FROM DUAL UNION ALL
SELECT 12, '02' FROM DUAL UNION ALL
SELECT 12, '03' FROM DUAL UNION ALL
SELECT 13, '04' FROM DUAL UNION ALL
SELECT 13, '01' FROM DUAL UNION ALL
SELECT 14, '02' FROM DUAL UNION ALL
SELECT 14, '01' FROM DUAL)
Expected output:
DAY
PEOPLE_LOGGED_EVERYDAY
PEOPLE
EXPLANATION
10
3
01, 02, 04
Three unique people in day 10
12
2
01, 02
Day 11 does not have values, so it's not included. From those in day 10, only 2 appear in day 12
13
1
01
From those in day 10 and 12, only 01 appears in day 13
14
1
01
From those in day 10, 12 and 13, only 01 appears in day 14
You can use listagg() with group by clause. If day is always start from the 1 and increases by 1 then you can use below query. He with the help of exits I have selected only those person_id which are available in all the previous days.
create table yourtable(DAY int, PERSON_ID varchar(10));
insert into yourtable values(1, '01');
insert into yourtable values(1, '02');
insert into yourtable values(1, '03');
insert into yourtable values(1, '04');
insert into yourtable values(2, '01');
insert into yourtable values(2, '02');
insert into yourtable values(2, '03');
insert into yourtable values(3, '01');
insert into yourtable values(4, '02');
insert into yourtable values(4, '01');
Query:
select day, count(person_id) as PEOPLE_LOGGED_EVERYDAY, LISTAGG(person_id,',') WITHIN GROUP(ORDER BY person_id) AS PEOPLE
from yourtable a
where exists (select 1 from yourtable b where b.day<=a.day and a.person_id=b.person_id
group by person_id having count(day)=a.day)
group by day;
Output:
DAY
PEOPLE_LOGGED_EVERYDAY
PEOPLE
1
4
01,02,03,04
2
3
01,02,03
3
1
01
4
1
01
db<fiddle here
Instead of day sequence if you had increasing dates in day column:
create table yourtable(DAY date, PERSON_ID varchar(10));
insert into yourtable values(date '2021-01-01', '01');
insert into yourtable values(date '2021-01-01', '02');
insert into yourtable values(date '2021-01-01', '03');
insert into yourtable values(date '2021-01-01', '04');
insert into yourtable values(date '2021-01-02', '01');
insert into yourtable values(date '2021-01-02', '02');
insert into yourtable values(date '2021-01-02', '03');
insert into yourtable values(date '2021-01-03', '01');
insert into yourtable values(date '2021-01-04', '02');
insert into yourtable values(date '2021-01-04', '01');
Query:
select day, count(person_id) as PEOPLE_LOGGED_EVERYDAY, LISTAGG(person_id,',') WITHIN GROUP(ORDER BY person_id) AS PEOPLE
from yourtable a
where exists (select 1 from yourtable b where b.day<=a.day and a.person_id=b.person_id
group by person_id having count(day)=( max(day)- min(day))+1)
group by day;
Output:
DAY
PEOPLE_LOGGED_EVERYDAY
PEOPLE
01-JAN-21
4
01,02,03,04
02-JAN-21
3
01,02,03
03-JAN-21
1
01
04-JAN-21
1
01
db<fiddle here
Revised answer
create table yourtable(DAY int, PERSON_ID varchar(10));
insert into yourtable(day,person_id)
with cte ( DAY, PERSON_ID ) AS(
SELECT 10, '01' FROM DUAL UNION ALL
SELECT 10, '02' FROM DUAL UNION ALL
SELECT 10, '04' FROM DUAL UNION ALL
SELECT 10, '04' FROM DUAL UNION ALL
SELECT 12, '01' FROM DUAL UNION ALL
SELECT 12, '02' FROM DUAL UNION ALL
SELECT 12, '03' FROM DUAL UNION ALL
SELECT 13, '04' FROM DUAL UNION ALL
SELECT 13, '01' FROM DUAL UNION ALL
SELECT 14, '02' FROM DUAL UNION ALL
SELECT 14, '01' FROM DUAL)
select * from cte ;
Query#1 (for Oracle 19c and later)
select day, count(person_id) as PEOPLE_LOGGED_EVERYDAY, LISTAGG(distinct person_id,',') WITHIN GROUP(ORDER BY person_id) AS PEOPLE
from yourtable a
where exists (select 1 from yourtable b where b.day<=a.day and a.person_id=b.person_id
group by person_id having count(DISTINCT day)=(select COUNT( distinct DAY) from yourtable where day<=a.day))
group by day;
Query#1 (for Oracle 18c and earlier)
select day, count(person_id) as PEOPLE_LOGGED_EVERYDAY, LISTAGG( person_id,',') WITHIN GROUP(ORDER BY person_id) AS PEOPLE
from
(
select distinct day, person_id
from yourtable a
where exists (select 1 from yourtable b where b.day<=a.day and a.person_id=b.person_id
group by person_id having count(DISTINCT day)=(select COUNT( distinct DAY) from yourtable where day<=a.day))
)t group by day
Output:
DAY
PEOPLE_LOGGED_EVERYDAY
PEOPLE
10
3
01,02,04
12
2
01,02
13
1
01
14
1
01
db<fiddle here
In Standard SQL, I would approach this by doing the following:
Enumerate the days for each person.
Determine the earliest day for each person.
Filter where the earliest day is "1" and the enumeration equals the days.
Then aggregate:
select day, count(*),
listagg(person_id, ',') within group (order by person_id)
from (select t.*,
row_number() over (partition by person_id order by day) as seqnum,
min(day) over (partition by person_id) as min_day
from t
) t
where seqnum = day and min_day = 1
group by day
order by day;
Note only is this simpler than using match recognize, but I would guess that the performance would be much better too.
You can use either:
SELECT DAY,
COUNT(DISTINCT person_id) AS num_people
FROM (
SELECT t.*,
DENSE_RANK() OVER (ORDER BY day)
- DENSE_RANK() OVER (PARTITION BY person_id ORDER BY day) AS day_grp
FROM table_name t
)
WHERE day_grp = 0
GROUP BY day
ORDER BY day
or MATCH_RECOGNIZE to find the successive days:
SELECT day,
COUNT(
DISTINCT
CASE cls WHEN 'CONSECUTIVE_DAYS' THEN person_id END
) AS num_people
FROM (
SELECT t.*,
DENSE_RANK() OVER (ORDER BY day) AS day_rank
FROM table_name t
)
MATCH_RECOGNIZE(
PARTITION BY person_id
ORDER BY day
MEASURES
classifier() AS cls
ALL ROWS PER MATCH
PATTERN ( ^ consecutive_days* )
DEFINE
consecutive_days AS COALESCE( PREV(day_rank) + 1, 1 ) = day_rank
)
GROUP BY day
ORDER BY day
Which, for the sample data:
CREATE TABLE table_name ( DAY, PERSON_ID ) AS
SELECT 1, '01' FROM DUAL UNION ALL
SELECT 1, '02' FROM DUAL UNION ALL
SELECT 1, '03' FROM DUAL UNION ALL
SELECT 1, '04' FROM DUAL UNION ALL
SELECT 2, '01' FROM DUAL UNION ALL
SELECT 2, '02' FROM DUAL UNION ALL
SELECT 2, '03' FROM DUAL UNION ALL
SELECT 3, '01' FROM DUAL UNION ALL
SELECT 3, '02' FROM DUAL UNION ALL
SELECT 4, '01' FROM DUAL;
Outputs:
DAY
NUM_PEOPLE
1
4
2
3
3
2
4
1
and for the sample data:
CREATE TABLE table_name ( DAY, PERSON_ID ) AS
SELECT 10, '01' FROM DUAL UNION ALL
SELECT 10, '02' FROM DUAL UNION ALL
SELECT 10, '04' FROM DUAL UNION ALL
SELECT 10, '04' FROM DUAL UNION ALL
SELECT 12, '01' FROM DUAL UNION ALL
SELECT 12, '02' FROM DUAL UNION ALL
SELECT 12, '03' FROM DUAL UNION ALL
SELECT 13, '04' FROM DUAL UNION ALL
SELECT 13, '01' FROM DUAL UNION ALL
SELECT 14, '02' FROM DUAL UNION ALL
SELECT 14, '01' FROM DUAL
Outputs:
DAY
NUM_PEOPLE
10
3
12
2
13
1
14
1
db<>fiddle here
Need to find record having gap between months in a table if the data is present in two different year.
I have column like id, value,month, year.
Id, value, month,year
1, 123, oct, 2020
1, 128, nov, 2020
1, 127, jan ,2021
2, 121, Dec, 2020
2, 154, jan, 2021
Output I need:
Id 1 as there is a gap in month (Dec is Missing for id=1)
Here's one option. Read comments within code.
SQL> with test (id, value, month, year) as
2 -- sample data; you have that, don't type it
3 (select 1, 123, 'oct', 2020 from dual union all
4 select 1, 128, 'nov', 2020 from dual union all
5 select 1, 127, 'jan', 2021 from dual union all
6 select 2, 121, 'dec', 2020 from dual union all
7 select 2, 154, 'jan', 2021 from dual
8 ),
9 temp as
10 -- "convert" month and year to real date value
11 (select id,
12 value,
13 to_date(month ||' '|| year, 'mon yyyy', 'nls_date_language=english') datum
14 from test
15 ),
16 temp2 as
17 -- select difference in months between DATUM and next month (LEAD!)
18 (select id,
19 months_between
20 (datum,
21 to_date(month ||' '|| year, 'mon yyyy', 'nls_date_language=english') datum
22 ) diff
23 from temp
24 )
25 select distinct id
26 from temp2
27 where abs(diff) > 1;
ID
----------
1
SQL>
It can probably be compressed, but step-by-step CTEs show what's going on.
I would construct a date and use lag():
select t.*
from (select t.*,
lag(dte) over (partition by id order by dte) as prev_dte
from (select t.*,
to_date(year || '-' || month || '-01', 'YYYY-MON-DD') as dte
from t
) t
) t
where prev_dte <> dte - interval '1' month;
Here is a db<>fiddle.
Here is an example using the LAG function and finding rows where where the prior month is not one month behind (or non existent)
WITH
sample_data (Id,
VALUE,
month,
year)
AS
(SELECT 1, 123, 'oct', 2020 FROM DUAL
UNION ALL
SELECT 1, 128, 'nov', 2020 FROM DUAL
UNION ALL
SELECT 1, 127, 'jan', 2021 FROM DUAL
UNION ALL
SELECT 2, 121, 'Dec', 2020 FROM DUAL
UNION ALL
SELECT 2, 154, 'jan', 2021 FROM DUAL)
SELECT DISTINCT id
FROM (SELECT sd.id,
CASE
WHEN ADD_MONTHS (TO_DATE (sd.year || sd.month, 'YYYYMON'), -1) =
TO_DATE (
LAG (sd.year || sd.month)
OVER (
PARTITION BY id
ORDER BY
sd.year, EXTRACT (MONTH FROM TO_DATE (sd.month, 'MON'))),
'YYYYMON')
OR LAG (sd.id)
OVER (
PARTITION BY id
ORDER BY sd.year, EXTRACT (MONTH FROM TO_DATE (sd.month, 'MON')))
IS NULL
THEN
'Y'
ELSE
'N'
END AS valid_prev_month
FROM sample_data sd)
WHERE valid_prev_month = 'N';
My table has records as below for different Id's and different start and end dates
ID, Startdate, Enddate
1, 2017-02-14, 2018-11-05
I want to write an SQL without using date dimension table that gives below output: Basically one record for each month between start and end date.
1, 2017, 02
1, 2017, 03
1, 2017, 04
1, 2017, 05
1, 2017, 06
1, 2017, 07
1, 2017, 08
1, 2017, 09
1, 2017, 10
1, 2017, 11
1, 2017, 12
1, 2018, 01
1, 2018, 02
1, 2018, 03
1, 2018, 04
1, 2018, 05
1, 2018, 06
1, 2018, 07
1, 2018, 09
1, 2018, 10
1, 2018, 11
Please use below query example:
set #start_date = '2017-02-14';
set #end_date = LAST_DAY('2018-11-05');
WITH RECURSIVE date_range AS
(
select MONTH(#start_date) as month_, YEAR(#start_date) as year_, DATE_ADD(#start_date, INTERVAL 1 MONTH) as next_month_date
UNION
SELECT MONTH(dr.next_month_date) as month_, YEAR(dr.next_month_date) as year_, DATE_ADD(dr.next_month_date, INTERVAL 1 MONTH) as next_month_date
FROM date_range dr
where next_month_date <= #end_date
)
select month_, year_ from date_range
order by next_month_date desc
This is what I did and it worked like a charm:
-- sample data
WITH table_data
AS (
SELECT 1 AS id
,cast('2017-08-14' AS DATE) AS start_dt
,cast('2018-12-16' AS DATE) AS end_dt
UNION ALL
SELECT 2 AS id
,cast('2017-09-14' AS DATE) AS start_dt
,cast('2019-01-16' AS DATE) AS end_dt
)
-- find minimum date from the data
,starting_date (start_date)
AS (
SELECT min(start_dt)
FROM TABLE_DATA
)
--get all months between min and max dates
,all_dates
AS (
SELECT last_day(add_months(date_trunc('month', start_date), idx * 1)) month_date
FROM starting_date
CROSS JOIN _v_vector_idx
WHERE month_date <= add_months(start_date, abs(months_between((
SELECT min(start_dt) FROM TABLE_DATA), (SELECT max(end_dt) FROM TABLE_DATA))) + 1)
ORDER BY month_date
)
SELECT id
,extract(year FROM month_date)
,extract(month FROM month_date)
,td.start_dt
,td.end_dt
FROM table_data td
INNER JOIN all_dates ad
ON ad.month_date > td.start_dt
AND ad.month_date <= last_day(td.end_dt)
ORDER BY 1
,2
You have to generate date and from that have to pick year and month
select distinct year(date),month( date) from
(select * from (
select
date_add('2017-02-14 00:00:00.000', INTERVAL n5.num*10000+n4.num*1000+n3.num*100+n2.num*10+n1.num DAY ) as date
from
(select 0 as num
union all select 1
union all select 2
union all select 3
union all select 4
union all select 5
union all select 6
union all select 7
union all select 8
union all select 9) n1,
(select 0 as num
union all select 1
union all select 2
union all select 3
union all select 4
union all select 5
union all select 6
union all select 7
union all select 8
union all select 9) n2,
(select 0 as num
union all select 1
union all select 2
union all select 3
union all select 4
union all select 5
union all select 6
union all select 7
union all select 8
union all select 9) n3,
(select 0 as num
union all select 1
union all select 2
union all select 3
union all select 4
union all select 5
union all select 6
union all select 7
union all select 8
union all select 9) n4,
(select 0 as num
union all select 1
union all select 2
union all select 3
union all select 4
union all select 5
union all select 6
union all select 7
union all select 8
union all select 9) n5
) a
where date >'2017-02-14 00:00:00.000' and date < '2018-11-05'
) as t
I'm trying to create a sql query which returns the first date of the current group.
Let's assume (just as an example) that on the first of every month a row is saved in a table (EmployeeInfo) with the employee ID and the current department.
EmployeeID, Department, Date (Format: DD.MM.YYYY)
100, IT, 01.07.2014
100, IT, 01.08.2014
100, IT, 01.09.2014
100, HR, 01.10.2014
100, HR, 01.11.2014
100, CC, 01.12.2014
100, IT, 01.01.2015
100, IT, 01.02.2015
100, IT, 01.03.2015
100, IT, 01.04.2015
The query should return the date since an employee is working in the current department.
The current department of the employee with ID 100 is IT, therefore the value should be 01.01.2015 (not 01.07.2014).
Any ideas how this could be implemented?
UPDATE New answer based on OP's comments.
You could use ANALYTIC function ROW_NUMBER and LAG.Something like, start of group method:
SQL> WITH DATA AS(
2 SELECT 100 EmployeeID, 'IT' Department, to_date('01.07.2014','DD.MM.YYYY') dt FROM dual UNION ALL
3 SELECT 100, 'IT', to_date('01.08.2014','DD.MM.YYYY') dt from dual union all
4 select 100, 'IT', to_date('01.09.2014','DD.MM.YYYY') dt from dual union all
5 SELECT 100, 'HR', to_date('01.10.2014','DD.MM.YYYY') dt from dual union all
6 select 100, 'HR', to_date('01.11.2014','DD.MM.YYYY') dt from dual union all
7 SELECT 100, 'CC', to_date('01.12.2014','DD.MM.YYYY') dt from dual union all
8 select 100, 'IT', to_date('01.01.2015','DD.MM.YYYY') dt from dual union all
9 select 100, 'IT', to_date('01.02.2015','DD.MM.YYYY') dt from dual
10 )
11 SELECT EmployeeID,
12 Department,
13 DT
14 FROM
15 (SELECT *
16 FROM
17 (SELECT t.*,
18 CASE
19 WHEN Department = lag(Department) over (PARTITION BY EmployeeID ORDER BY dt)
20 THEN 0
21 ELSE 1
22 END gap
23 FROM DATA t
24 ) T
25 WHERE GAP = 1
26 ORDER BY DT DESC
27 )
28 WHERE ROWNUM = 1
29 /
EMPLOYEEID DE DT
---------- -- ---------
100 IT 01-JAN-15
SQL>
OLD answer
For example,
SQL> WITH DATA AS(
2 SELECT 100 EmployeeID, 'IT' Department, to_date('01.07.2014','DD.MM.YYYY') dt FROM dual UNION ALL
3 SELECT 100, 'IT', to_date('01.08.2014','DD.MM.YYYY') dt from dual union all
4 select 100, 'IT', to_date('01.09.2014','DD.MM.YYYY') dt from dual union all
5 SELECT 100, 'HR', to_date('01.10.2014','DD.MM.YYYY') dt from dual union all
6 select 100, 'HR', to_date('01.11.2014','DD.MM.YYYY') dt from dual union all
7 SELECT 100, 'CC', to_date('01.12.2014','DD.MM.YYYY') dt from dual union all
8 select 100, 'IT', to_date('01.01.2015','DD.MM.YYYY') dt from dual union all
9 select 100, 'IT', to_date('01.02.2015','DD.MM.YYYY') dt from dual
10 )
11 SELECT*
12 FROM
13 (SELECT t.*,
14 row_number() OVER(PARTITION BY department ORDER BY dt DESC) rn
15 FROM DATA t
16 )
17 WHERE rn = 1
18 /
EMPLOYEEID DE DT RN
---------- -- --------- ----------
100 CC 01-DEC-14 1
100 HR 01-NOV-14 1
100 IT 01-FEB-15 1
SQL>
Use below query and let me know if want this
WITH table_ (EmployeeID, Department, Dat) AS (
SELECT 100, 'IT', TO_DATE('01.07.2014', 'DD.MM.YYYY') FROM DUAL UNION ALL
SELECT 100, 'IT', TO_DATE('01.08.2014', 'DD.MM.YYYY') FROM DUAL UNION ALL
SELECT 100, 'IT', TO_DATE('01.09.2014', 'DD.MM.YYYY') FROM DUAL UNION ALL
SELECT 100, 'HR', TO_DATE('01.10.2014', 'DD.MM.YYYY') FROM DUAL UNION ALL
SELECT 100, 'HR', TO_DATE('01.11.2014', 'DD.MM.YYYY') FROM DUAL UNION ALL
SELECT 100, 'CC', TO_DATE('01.12.2014', 'DD.MM.YYYY') FROM DUAL UNION ALL
SELECT 100, 'IT', TO_DATE('01.01.2015', 'DD.MM.YYYY') FROM DUAL UNION ALL
SELECT 100, 'IT', TO_DATE('01.02.2015', 'DD.MM.YYYY') FROM DUAL),
----------
--End of Data preparation
----------
table1 AS (SELECT EmployeeID, Department, Dat,
row_number() OVER (PARTITION BY EmployeeID ORDER BY dat DESC) -
row_number() OVER (PARTITION BY EmployeeID, department ORDER BY dat DESC) rk
FROM TABLE_),
table2 AS (SELECT EmployeeID, Department, Dat, RANK() OVER (PARTITION BY employeeid ORDER BY dat) rnk
from table1
WHERE rk = 0 )
SELECT EmployeeID, Department, Dat
FROM table2
WHERE rnk = 1;
output:
EMPLOYEEID DE DAT
---------- -- ---------
100 IT 01-JAN-15