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);
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
)
This question already has answers here:
How to get the previous working day from Oracle?
(4 answers)
Closed 1 year ago.
need help for some oracle stuff ..
I need to get Day-1 from sysdate, holiday and weekend will be excluded .
And for holiday, we need to get the range to get the last workday before holiday.
The start date and end date will coming from my holiday table.
ex :
Holiday Table
HolidayName
Start_date
End_Date
holiday1
5th Aug'21
6th Aug'21
condition :
this query run on 9th Aug 2021
expected result :
4th Aug'21
I've tried some query and function but I just can't get what I need.
Thanks a lot for help!
Here's one way to do it.
select max(d) as last_workday
from (select trunc(sysdate)-level as d from dual connect by level < 30) prior_month
where to_char(d, 'DY') not in ('SAT','SUN')
and not exists (select holidayname from holiday_table
where prior_month.d between start_date and end_date)
;
Without seeing your Holiday table, it's hard to say how many days back you would need to look to find the last workday. If you have a holiday that lasts for more than 30 days, you'll need to change the 30 to a larger number.
You can use a simple case expression to determine what day of the week the start of your holiday is, then subtract a number of days based on that.
WITH
holiday (holidayname, start_date, end_date)
AS
(SELECT 'holiday1', DATE '2021-8-5', DATE '2021-8-6' FROM DUAL
UNION ALL
SELECT 'Christmas', DATE '2021-12-25', DATE '2021-12-26' FROM DUAL
UNION ALL
SELECT 'July 4th', DATE '2021-7-4', DATE '2021-7-5' FROM DUAL)
SELECT holidayname,
start_date,
end_date,
start_date - CASE TO_CHAR (start_date, 'Dy') WHEN 'Mon' THEN 3 WHEN 'Sun' THEN 2 ELSE 1 END AS prior_business_day
FROM holiday;
HOLIDAYNAME START_DATE END_DATE PRIOR_BUSINESS_DAY
______________ _____________ ____________ _____________________
holiday1 05-AUG-21 06-AUG-21 04-AUG-21
Christmas 25-DEC-21 26-DEC-21 24-DEC-21
July 4th 04-JUL-21 05-JUL-21 02-JUL-21
You can use a recursive sub-query factoring clause from this answer:
WITH start_date (dt) AS (
SELECT DATE '2021-05-02' FROM DUAL
),
days ( dt, day, found ) AS (
SELECT dt,
TRUNC(dt) - TRUNC(dt, 'IW'),
0
FROM start_date
UNION ALL
SELECT dt - CASE day WHEN 0 THEN 3 WHEN 6 THEN 2 ELSE 1 END,
CASE WHEN day IN (0, 6, 5) THEN 4 ELSE day - 1 END,
CASE WHEN h.start_date IS NULL THEN 1 ELSE 0 END
FROM days d
LEFT OUTER JOIN holidays h
ON ( dt - CASE day WHEN 0 THEN 3 WHEN 6 THEN 2 ELSE 1 END
BETWEEN h.start_date AND h.end_date )
WHERE found = 0
)
SELECT dt
FROM days
WHERE found = 1;
Which, for the sample data:
CREATE TABLE holidays (HolidayName, Start_date, End_Date) AS
SELECT 'holiday1', DATE '2021-08-05', DATE '2021-08-06' FROM DUAL;
Outputs:
DT
2021-08-04 00:00:00
db<>fiddle here
Don't know if it's very efficient. Did it just for fun
create table holidays (
holiday_name varchar2(100) primary key,
start_date date not null,
end_date date not null
)
/
Table created
insert into holidays (holiday_name, start_date, end_date)
values ('holiday1', date '2021-08-05', date '2021-08-06');
1 row inserted
with days_before(day, wrk_day) as
(select trunc(sysdate - 1) d,
case
when h.holiday_name is not null then 0
when to_char(trunc(sysdate - 1), 'D') in ('6', '7') then 0
else 1
end work_day
from dual
left join holidays h
on trunc(sysdate - 1) between h.start_date and h.end_date
union all
select db.day - 1,
case
when h.holiday_name is not null then 0
when to_char(db.day - 1, 'D') in ('6', '7') then 0
else 1
end work_day
from days_before db
left join holidays h
on db.day - 1 between h.start_date and h.end_date
where db.wrk_day = 0) search depth first by day set order_no
select day from days_before where wrk_day = 1;
DAY
-----------
04.08.2021
I have a table with some fileds like these:
**Date Name Table Direction Code**
13/01/1978 Jacks xxxx stret ... 1
13/01/1978 John xxxx xxxxx 0
...........
12/01/1978 Dave xxxx xxxxxx 0
12/01/1978 Suse xxxxx xxxxxx 0
...........
11/01/1978 Mickey xxxx xxxxx 1
11/01/1978 Suse xxxxxx xxxxxxx 2
11/01/1978 Nune xxxxxx xxxxxxx 2
.....
09/01/1978 ..... ...... ....... 1
08/01/1978 ..... ...... ....... 0
21/01/1978 ..... ...... ....... 1
.....
I would like to extract the three first ocurrences grouping by date descendent and detect if the field code have an value '1' in any row with the same date. The result that I would like is:
The first date is 21/01/1978, and have code 1
The second date is 13/01/1978 and have code 1
The third date is 12/01/1978 and don't have code 1
I'm trying with rownum but not works. Could you help me please? Thanks and sorry for my English!
I don't know why rownum didn't work for u, maybe becuase you tried to put in in the inner query before the order by(you first have to order and only then use rownum becuase every row will get another rownum)
SELECT Date,code from (
(SELECT Date,max(Code) as code from YourTable
GROUP BY Date
ORDER BY Date DESC
)
WHERE rownum <= 3
Try this Hope it helps
SELECT DISTINCT B.DT,
DECODE(B.VL,1,' has code 1',999,' dont have code 1') inf
FROM
(SELECT A.DT,
CASE
WHEN A.CODE = 1
THEN 1
ELSE 999
END VL,
ROW_NUMBER() OVER(PARTITION BY A.DT ORDER BY A.CODE DESC) RN
FROM
(SELECT '01/03/2016' AS DT,1 AS CODE FROM DUAL
UNION ALL
SELECT '01/03/2016' AS DT,0 AS CODE FROM DUAL
UNION ALL
SELECT '05/03/2016' AS DT,0 AS CODE FROM DUAL
UNION ALL
SELECT '05/03/2016' AS DT,0 AS CODE FROM DUAL
UNION ALL
SELECT '06/03/2016' AS DT,1 AS CODE FROM DUAL
)A
) B
WHERE B.RN = 1;
you can do it in quite a few ways, one of them would be like:
select date,
decode(date_cnt, 0, 'dont have code = 1', 'have code = 1)
from (
select t.date,
max(t.code) as date_cnt
from table_name t
group by date
order by date desc)
where rownum <=3
this way you first sort them by date in the inner query aswell as prepare the info about having a code = 1. The outer query will fetch first 3 rows and change the codes to desired info (if 0 => no code, else there is code)
Hi I am working on oracle DB. The DB has one column date. It consists of the dates of 5 years with the dates model will be refreshed. Ex
DATE_TABLE
DATE
------------
1-jan-2013
15-jan-2013
31-jan-2013
6-feb-2013
etc.........
now for today's date suppose 13th jan 2013. The next refresh date will be 15th jan. and previous refresh date is 1st jan. to retrieve these two dates. Can i have any way without using PL/SQL. using regular select queries?. Thanks in advance
There are two functions LAG() (allows you to reference previous record) and LEAD() allows you to reference next record. Here is an example:
SQL> with t1(col) as(
2 select '1-jan-2013' from dual union all
3 select '15-jan-2013' from dual union all
4 select '31-jan-2013' from dual union all
5 select '6-feb-2013' from dual
6 )
7 select col as current_value
8 , lag(col, 1) over(order by col) as prev_value
9 , lead(col, 1) over(order by col)as next_value
10 from t1
11 ;
Result:
CURRENT_VALUE PREV_VALUE NEXT_VALUE
------------- ----------- -----------
1-jan-2013 NULL 15-jan-2013
15-jan-2013 1-jan-2013 31-jan-2013
31-jan-2013 15-jan-2013 6-feb-2013
6-feb-2013 31-jan-2013 NULL
We can simply use the below query, plain and simple. No need of pl/sql
SELECT MIN(DATE) FROM DATE_TABLE WHERE DATE > SYSDATE ;
i have below data.
table A
id
1
2
3
table B
id name data1 data2 datetime
1 cash 12345.00 12/12/2012 11:10:12
1 quantity 222.12 14/12/2012 11:10:12
1 date 20/12/2012 12/12/2012 11:10:12
1 date 19/12/2012 13/12/2012 11:10:12
1 date 13/12/2012 14/12/2012 11:10:12
1 quantity 330.10 17/12/2012 11:10:12
I want to retrieve data in one row like below:
tableA.id tableB.cash tableB.date tableB.quantity
1 12345.00 13/12/2012 330.10
I want to retrieve based on max(datetime).
The data model appears to be insane-- it makes no sense to join an ORDER_ID to a CUSTOMER_ID. It makes no sense to store dates in a VARCHAR2 column. It makes no sense to have no relationship between a CUSTOMER and an ORDER. It makes no sense to have two rows in the ORDER table with the same ORDER_ID. ORDER is also a reserved word so you cannot use that as a table name. My best guess is that you want something like
select *
from customer c
join (select order_id,
rank() over (partition by order_id
order by to_date( order_time, 'YYYYMMDD HH24:MI:SS' ) desc ) rnk
from order) o on (c.customer_id=o.order_id)
where o.rnk = 1
If that is not what you want, please (as I asked a few times in the comments) post the expected output.
These are the results I get with my query and your sample data (fixing the name of the ORDER table so that it is actually valid)
SQL> ed
Wrote file afiedt.buf
1 with orders as (
2 select 1 order_id, 'iphone' order_name, '20121201 12:20:23' order_time from dual union all
3 select 1, 'iphone', '20121201 12:22:23' from dual union all
4 select 2, 'nokia', '20110101 13:20:20' from dual ),
5 customer as (
6 select 1 customer_id, 'paul' customer_name from dual union all
7 select 2, 'stuart' from dual union all
8 select 3, 'mike' from dual
9 )
10 select *
11 from customer c
12 join (select order_id,
13 rank() over (partition by order_id
14 order by to_date( order_time, 'YYYYMMDD HH24:MI:SS' ) desc ) rnk
15 from orders) o on (c.customer_id=o.order_id)
16* where o.rnk = 1
SQL> /
CUSTOMER_ID CUSTOM ORDER_ID RNK
----------- ------ ---------- ----------
1 paul 1 1
2 stuart 2 1
Try something like
SELECT *
FROM CUSTOMER c
INNER JOIN ORDER o
ON (o.CUSTOMER_ID = c.CUSTOMER_ID)
WHERE TO_DATE(o.ORDER_TIME, 'YYYYMMDD HH24:MI:SS') =
(SELECT MAX(TO_DATE(o.ORDER_TIME, 'YYYYMMDD HH24:MI:SS')) FROM ORDER)
Share and enjoy.