How to find the distinct dates in a year using Oracle?
id sent_date
1 2017-05-01
1 2017-05-01
1 2017-06-01
1 2016-06-01
Ignore the duplicate sent_date for a Id in the same year.
Output
count(*) id year
2 1 2017
1 1 2016
EDIT:
Thisis my query
select distinct(count(sent_date)), id , extract (year from sent_date)
from test
GROUP BY id, extract (year from sent_date).
3 1 2017 (wrong)-- expecting the count as 2 1 1 2016 –
The DISTINCT is wrongly positioned in your query; you simply need:
select count(distinct sent_date), id , extract (year from sent_date)
from test
group by id, extract (year from sent_date)
Also, DISTINCT is not a function, so the syntax DISTINCT(...) does not make sense.
Related
I have a very big table 'DATES_EVENTS' (20 T) that looks like this:
ID DATE
1 '2022-04-01'
1 '2022-03-02'
1 '2022-03-01'
2 '2022-05-01'
3 '2021-12-01'
3 '2021-11-11'
3 '2020-11-11'
3 '2020-10-01'
I want per each row to get all past dates (per user) limited to up to 6 months.
My desired table:
ID DATE DATE_list
1 '2022-04-01' ['2022-04-01','2022-03-02','2022-03-01']
1 '2022-03-02' ['2022-03-02','2022-03-01']
1 '2022-03-01' ['2022-03-01']
2 '2022-05-01' ['2022-05-01']
3 '2021-12-01' ['2021-12-01','2021-11-11']
3 '2021-11-11' ['2021-11-11']
3 '2020-11-11' ['2020-11-11','2020-10-01']
3 '2020-10-01' ['2020-10-01']
I have a solution for all dates not limited:
SELECT
ID, DATE, ARRAY_AGG(DATE) OVER (PARTITION BY ID ORDER BY DATE) as DATE_list
FROM
DATES_EVENTS
But for a limited up to 6 months I don't have an efficient solution:
SELECT
distinct A.ID, A.DATE, ARRAY_AGG(B.DATE) OVER (PARTITION BY B.ID ORDER BY B.DATE) as DATE_list
FROM
DATES_EVENTS A
INNER JOIN
DATES_EVENTS B
ON
A.ID=B.ID
AND B.DATE BETWEEN DATE_SUB(A.DATE, INTERVAL 180 DAY) AND A.DATE
** ruffly a solution
Anyone know of a good and efficient way to do what I need?
Consider below approach
select id, date, array(
select day
from t.date_list day
where day <= date
order by day desc
) as date_list
from (
select *, array_agg(date) over win as date_list
from dates_events
window win as (
partition by id
order by extract(year from date) * 12 + extract(month from date)
range between 5 preceding and current row
)
) t
if applied to sample data in your question - output is
In case if (as I noticed in your question) 180 days is appropriate substitution for 6 months for you - you can use below simpler version
select *, array_agg(date) over win as date_list
from dates_events
window win as (
partition by id
order by unix_date(date)
range between current row and 179 following
)
I don't write too many queries. So, it might be a simple question for you. I need to rank a set based on a date in desc order and by a sequence number in asc order. For example:
ID
Date
Seq
Rank
1
Aug 1
0
1
1
Aug 1
1
2
1
Jul 1
0
3
1
Jun 1
0
4
1
May 1
0
5
How would I achieve that? The Rank would only allow you to order by multiple fields but by just either asc or desc but not both, right?
Thank you in advance
You can try the below query -
SELECT ID, "Date", Seq, ROW_NUMBER() OVER(ORDER BY TO_DATE("Date", 'Mon dd') DESC, Seq) Rank
FROM YOUR_TABLE;
I group my table by months
SELECT TO_CHAR (created, 'YYYY-MM') AS operation, COUNT (id)
FROM user_info
WHERE created IS NOT NULL
GROUP BY ROLLUP (TO_CHAR (created, 'YYYY-MM'))
2015-04 1
2015-06 10
2015-08 22
2015-09 8
2015-10 13
2015-12 5
2016-01 25
2016-02 37
2016-03 24
2016-04 1
2016-05 1
2016-06 2
2016-08 2
2016-09 7
2016-10 103
2016-11 5
2016-12 2
2017-04 14
2017-05 2
284
But the records don't cover all the months.
I would like the output to include all the months, with the missing ones displayed in the output with a default value:
2017-01 ...
2017-02 ...
2017-03 ZERO
2017-04 ZERO
2017-05 ...
Oracle has a good array of date manipulation functions. The two pertinent ones for this problem are
MONTHS_BETWEEN() which calculates the number of months between two dates
ADD_MONTHS() which increments a date by the given number of months
We can combine these functions to generate a table of all the months spanned by your table's records. Then we use an outer join to conditionally join records from USER_INFO to that calendar. When no records match count(id) will be zero.
with cte as (
select max(trunc(created, 'MM')) as max_dt
, min(trunc(created, 'MM')) as min_dt
from user_info
)
, cal as (
select add_months(min_dt, (level-1)) as mth
from cte
connect by level <= months_between(max_dt, min_dt) + 1
)
select to_char(cal.mth, 'YYYY-MM') as operation
, count(id)
from cal
left outer join user_info
on trunc(user_info.created, 'mm') = cal.mth
group by rollup (cal.mth)
order by 1
/
I have a problem which can be handled by a recursive CTE, but not within an acceptable period of time. Can anyone point me at ways to improve the performance and/or get the same result a different way?
Here's my scenario!
I have : A large table which contains in each row an id, a start date, an end date, and a ranking number. There are multiple rows for each id and the date ranges often overlap. Dates are from 2010 onward.
I want: A table which contains a row for each combination of id + date which falls inside any date range for that id from the previous table. Each row should have the lowest ranking number for that id and day.
Eg:
ID Rank Range
1 1 1/1/2010-1/4/2010
1 2 1/2/2010-1/5/2010
2 1 1/1/2010-1/2/2010
becomes
ID Rank Day
1 1 1/1/2010
1 1 1/2/2010
1 1 1/3/2010
1 1 1/4/2010
1 2 1/5/2010
2 1 1/1/2010
2 1 1/2/2010
I can do this with a recursive CTE, but the performance is terrible (20-25 minutes for a relatively small data set which produces a final table with 31 million rows):
with enc(PersonID, EncounterDate, EndDate, Type_Rank) as (
select PersonID, EncounterDate, EndDate, Type_Rank
from Big_Base_Table
union all
select PersonID, EncounterDate + 1, EndDate, Type_Rank
from enc
where EncounterDate + 1 <= EndDate
)
select PersonID, EncounterDate, min(Type_Rank) Type_Rank
from enc
group by PersonID, EncounterDate
;
You could extract all possible dates from the table once in a CTE, and then join that back to the table:
with all_dates (day) as (
select start_date + level - 1
from (
select min(start_date) as start_date, max(end_date) as end_date
from big_base_table
)
connect by level <= end_date - start_date + 1
)
select bbt.id, min(bbt.type_rank) as type_rank, to_char(ad.day, 'YYYY-MM-DD') as day
from all_dates ad
join big_base_table bbt
on bbt.start_date <= ad.day
and bbt.end_date >= ad.day
group by bbt.id, ad.day
order by bbt.id, ad.day;
ID TYPE_RANK DAY
---------- ---------- ----------
1 1 2010-01-01
1 1 2010-01-02
1 1 2010-01-03
1 1 2010-01-04
1 2 2010-01-05
2 1 2010-01-01
2 1 2010-01-02
7 rows selected.
The CTE gets all dates from the lowest for any ID, up to the highest for any ID. You could also use a static calendar table for that if you have one, to save hitting the table twice (and getting min/max at the same time is slow in some versions at least).
You could also write it the other way round, as:
...
from big_base_table bbt
join all_dates ad
on ad.day >= bbt.start_date
and ad.day <= bbt.end_date
...
but I think the optimisier will probably end up treating them the same, with a single full scan of your base table; worth checking the plan it actually comes up with for both though, and if one is more efficnet that the other.
I have an Oracle table with data like below:
1. ID DATE
2. 12 02/11/2013
3. 12 02/12/2013
4. 13 02/11/2013
5. 13 02/12/2013
6. 13 02/13/2013
7. 13 02/14/2013
8. 14 02/11/2013
9. 14 02/12/2013
10. 14 02/13/2013
I need to find only those ID who has only Monday, Tuesday and Wednesday dates, so here only ID = 14 should be returned. I am using Oracle and dates are in format MM/DD/YYYY.
Please advice.
Regards,
Nitin
If date column is DATE datatype, then you can
select id
from your_table
group by id
having sum(case
when to_char(date_col,'fmday')
in ('monday','tuesday','wednesday') then 1
else 99
end) = 3;
EDIT: Corected the above code at the igr's observation
But this is ok only if you don't have a day twice for the same id.
If the column is varchar2 then the condition becomes to_char(to_date(your_col,'mm/dd/yyyy'),'fmday') in ...
A more robust code would be:
select id
from(
select id, date_col
from your_table
group by id, date_col
)
group by id
having sum(case
when to_char(date_col,'fmday', 'NLS_DATE_LANGUAGE=ENGLISH')
in ('monday','tuesday','wednesday') then 1
else 99
end) = 3;
select id
from (
select
id,
sum (case when to_char(dt, 'D', 'nls_territory=AMERICA') between 1 and 3 then 1 else -1 end) AS cnt
from t
group by id
)
where cnt=3
NOTE: I assumed (id,dt) is unique - no two lines with same id and date.
do something like
SELECT * FROM your_table t
where to_char(t.DATE, 'DY') in ('whatever_day_abbreviation_day_you_use');
alternatively if you prefer you could use day numbers like:
SELECT * FROM your_table t
where to_number(to_char(d.ts, 'D')) in (1,2,3);
if you'd like to avoid ID repetition add DISTINCTION
SELECT DISTINCT ID FROM your_table t
where to_number(to_char(d.ts, 'D')) in (1,2,3);