I'm trying to get the highest date and lowest date in the column, from there get the data that is in the filesize column of the rows that have these respective dates and add it to another column.
I made the following code but it doesn't work.
CASE WHEN MIN(to_timestamp(extract_dt, 'yyyy-MM-dd')) THEN filesize ELSE 0 END filesize_extracao_passada,
You can use a SQL like below.
select
filesize,dt,
if ( mx.mxdt = to_timestamp(extract_dt, 'yyyy-MM-dd'), filesize,null) mx_dt, -- this compares and display filesize for max dt
if ( mn.mndt = to_timestamp(extract_dt, 'yyyy-MM-dd'), filesize,null) mn_dt
from mytable
left join (select max(to_timestamp(extract_dt, 'yyyy-MM-dd')) mxdt from tmp)mx on 1=1 --This calculates maximum date
left join (select min(to_timestamp(extract_dt, 'yyyy-MM-dd')) mndt from tmp)mn on 1=1 --This calculates minimum date
WHERE
mn.mndt = to_timestamp(extract_dt, 'yyyy-MM-dd') OR mx.mxdt = to_timestamp(extract_dt, 'yyyy-MM-dd') -- this ensure your select clause returns only rows having maximum or minimum dates
Here is my code if you want to try out in your environment.
create table tmp as
select 'a' as id, now() dt union all
select 'a' as id, now()+ interval 1 days dt union all
select 'a' as id, now() - interval 5 days dt union all
select 'a' as id, now()+ interval 3 days dt union all
select 'a' as id, now()+ interval 3 days dt union all
select 'a' as id, now()+ interval 1 days dt union all
select 'a' as id, now()+ interval 2 days dt ;
select
id,dt,
if ( mx.mxdt = dt, id,null) mx_dt,
if ( mn.mndt = dt, id,null) mn_dt
from tmp
left join (select max(dt) mxdt from tmp)mx on 1=1
left join (select min(dt) mndt from tmp)mn on 1=1
WHERE mx.mxdt = dt or mn.mndt = dt
Related
I have three tabels, each of them has a date column (the date column is an INT field and needs to stay that way). I need a UNION accross all three tables so that I get the list of unique dates in accending order like this:
20040602
20051215
20060628
20100224
20100228
20100422
20100512
20100615
Then I need to add a column to the result of the query where I subtract one from each date and place it one row above as the end date. Basically I need to generate the end date from the start date somehow and this is what I got so far (not working):
With Query1 As (
Select date_one As StartDate
From table_one
Union
Select date_two As StartDate
From table_two
Union
Select date_three e As StartDate
From table_three
Order By Date Asc
)
Select Query1.StartDate - 1 As EndDate
From Query1
Thanks a lot for your help!
Building on your existing union cte, we can use lead() in the outer query to get the start_date of the next record, and withdraw 1 from it.
with q as (
select date_one start_date from table_one
union select date_two from table_two
union select date_three from table_three
)
select
start_date,
dateadd(day, -1, lead(start_date) over(order by start_date)) end_date
from q
order by start_date
If the datatype the original columns are numeric, then you need to do some casting before applying date functions:
with q as (
select cast(cast(date_one as varchar(8)) as date) start_date from table_one
union select cast(cast(date_two as varchar(8)) as date) from table_two
union select cast(cast(date_three as varchar(8)) as date) from table_three
)
select
start_date,
dateadd(day, -1, lead(start_date) over(order by start_date)) end_date
from q
order by start_date
I am working on a "counting days" problem almost identical to this one. I have a list of date(s), and need to count how many days used excluding duplicate, and handling the gaps. Same input and output.
From: Markus Jarderot
Input
ID d1 d2
1 2011-08-01 2011-08-08
1 2011-08-02 2011-08-06
1 2011-08-03 2011-08-10
1 2011-08-12 2011-08-14
2 2011-08-01 2011-08-03
2 2011-08-02 2011-08-06
2 2011-08-05 2011-08-09
Output
ID hold_days
1 11
2 8
SQL to find time elapsed from multiple overlapping intervals
But for the life of me I couldn't understand Markus Jarderot's solution.
SELECT DISTINCT
t1.ID,
t1.d1 AS date,
-DATEDIFF(DAY, (SELECT MIN(d1) FROM Orders), t1.d1) AS n
FROM Orders t1
LEFT JOIN Orders t2 -- Join for any events occurring while this
ON t2.ID = t1.ID -- is starting. If this is a start point,
AND t2.d1 <> t1.d1 -- it won't match anything, which is what
AND t1.d1 BETWEEN t2.d1 AND t2.d2 -- we want.
GROUP BY t1.ID, t1.d1, t1.d2
HAVING COUNT(t2.ID) = 0
Why is DATEDIFF(DAY, (SELECT MIN(d1) FROM Orders), t1.d1) picking from the min(d1) from the entire list? Is that regardless of ID.
And what does t1.d1 BETWEEN t2.d1 AND t2.d2 do? Is that to ensure only overlapped interval are calculated?
Same thing with group by, I think because if in the event the same identical period will be discarded? I tried to trace the solution by hand but getting more confused.
This is mostly a duplicate of my answer here (including explanation) but with the inclusion of grouping on an id column. It should use a single table scan and does not require a recursive sub-query factoring clause (CTE) or self joins.
SQL Fiddle
Oracle 11g R2 Schema Setup:
CREATE TABLE your_table ( id, usr, start_date, end_date ) AS
SELECT 1, 'A', DATE '2017-06-01', DATE '2017-06-03' FROM DUAL UNION ALL
SELECT 1, 'B', DATE '2017-06-02', DATE '2017-06-04' FROM DUAL UNION ALL -- Overlaps previous
SELECT 1, 'C', DATE '2017-06-06', DATE '2017-06-06' FROM DUAL UNION ALL
SELECT 1, 'D', DATE '2017-06-07', DATE '2017-06-07' FROM DUAL UNION ALL -- Adjacent to previous
SELECT 1, 'E', DATE '2017-06-11', DATE '2017-06-20' FROM DUAL UNION ALL
SELECT 1, 'F', DATE '2017-06-14', DATE '2017-06-15' FROM DUAL UNION ALL -- Within previous
SELECT 1, 'G', DATE '2017-06-22', DATE '2017-06-25' FROM DUAL UNION ALL
SELECT 1, 'H', DATE '2017-06-24', DATE '2017-06-28' FROM DUAL UNION ALL -- Overlaps previous and next
SELECT 1, 'I', DATE '2017-06-27', DATE '2017-06-30' FROM DUAL UNION ALL
SELECT 1, 'J', DATE '2017-06-27', DATE '2017-06-28' FROM DUAL UNION ALL -- Within H and I
SELECT 2, 'K', DATE '2011-08-01', DATE '2011-08-08' FROM DUAL UNION ALL -- Your data below
SELECT 2, 'L', DATE '2011-08-02', DATE '2011-08-06' FROM DUAL UNION ALL
SELECT 2, 'M', DATE '2011-08-03', DATE '2011-08-10' FROM DUAL UNION ALL
SELECT 2, 'N', DATE '2011-08-12', DATE '2011-08-14' FROM DUAL UNION ALL
SELECT 3, 'O', DATE '2011-08-01', DATE '2011-08-03' FROM DUAL UNION ALL
SELECT 3, 'P', DATE '2011-08-02', DATE '2011-08-06' FROM DUAL UNION ALL
SELECT 3, 'Q', DATE '2011-08-05', DATE '2011-08-09' FROM DUAL;
Query 1:
SELECT id,
SUM( days ) AS total_days
FROM (
SELECT id,
dt - LAG( dt ) OVER ( PARTITION BY id
ORDER BY dt ) + 1 AS days,
start_end
FROM (
SELECT id,
dt,
CASE SUM( value ) OVER ( PARTITION BY id
ORDER BY dt ASC, value DESC, ROWNUM ) * value
WHEN 1 THEN 'start'
WHEN 0 THEN 'end'
END AS start_end
FROM your_table
UNPIVOT ( dt FOR value IN ( start_date AS 1, end_date AS -1 ) )
)
WHERE start_end IS NOT NULL
)
WHERE start_end = 'end'
GROUP BY id
Results:
| ID | TOTAL_DAYS |
|----|------------|
| 1 | 25 |
| 2 | 13 |
| 3 | 9 |
The brute force method is to create all days (in a recursive query) and then count:
with dates(id, day, d2) as
(
select id, d1 as day, d2 from mytable
union all
select id, day + 1, d2 from dates where day < d2
)
select id, count(distinct day)
from dates
group by id
order by id;
Unfortunately there is a bug in some Oracle versions and recursive queries with dates don't work there. So try this code and see whether it works in your system. (I have Oracle 11.2 and the bug still exists there; so I guess you need Oracle 12c.)
I guess Markus' idea is to find all starting points that are not within other ranges and all ending points that aren't. Then just take the first starting point till the first ending point, then the next starting point till the next ending point, etc. As Markus isn't using a window function to number starting and ending points, he must find a more complicated way to achieve this. Here is the query with ROW_NUMBER. Maybe this gives you a start what to look for in Markus' query.
select startpoint.id, sum(endpoint.day - startpoint.day)
from
(
select id, d1 as day, row_number() over (partition by id order by d1) as rn
from mytable m1
where not exists
(
select *
from mytable m2
where m1.id = m2.id
and m1.d1 > m2.d1 and m1.d1 <= m2.d2
)
) startpoint
join
(
select id, d2 as day, row_number() over (partition by id order by d1) as rn
from mytable m1
where not exists
(
select *
from mytable m2
where m1.id = m2.id
and m1.d2 >= m2.d1 and m1.d2 < m2.d2
)
) endpoint on endpoint.id = startpoint.id and endpoint.rn = startpoint.rn
group by startpoint.id
order by startpoint.id;
If all your intervals start at different dates, consider them in ascending order by d1 counting how many days are from d1 to the next interval.
You can discard an interval of it is contained in another one.
The last interval won't have a follower.
This query should give you how many days each interval give
select a.id, a.d1,nvl(min(b.d1), a.d2) - a.d1
from orders a
left join orders b
on a.id = b.id and a.d1 < b.d1 and a.d2 between b.d1 and b.d2
group by a.id, a.d1
Then group by id and sum days
I am arriving at the total number of days a service has been used in a month.
(Start_Date and End_Date are - both inclusive)
Sample Data 1:
User Start_Date End_Date
A 01-Jun-2017 30-Jun-2017
B 06-Jun-2017 30-Jun-2017
Ans: Service used days = 30 days.
Sample Data 2:
User Start_Date End_Date
C 06-Jun-2017 10-Jun-2017
D 02-Jun-2017 02-Jun-2017
Ans: Service used days = 6 days.
How do I write a code to find the same, preferable in SQL to PLSQL.
Test Data:
CREATE TABLE your_table ( usr, start_date, end_date ) AS (
SELECT 'A', DATE '2017-06-01', DATE '2017-06-03' FROM DUAL UNION ALL
SELECT 'B', DATE '2017-06-02', DATE '2017-06-04' FROM DUAL UNION ALL -- Overlaps previous
SELECT 'C', DATE '2017-06-06', DATE '2017-06-06' FROM DUAL UNION ALL
SELECT 'D', DATE '2017-06-07', DATE '2017-06-07' FROM DUAL UNION ALL -- Adjacent to previous
SELECT 'E', DATE '2017-06-11', DATE '2017-06-20' FROM DUAL UNION ALL
SELECT 'F', DATE '2017-06-14', DATE '2017-06-15' FROM DUAL UNION ALL -- Within previous
SELECT 'G', DATE '2017-06-22', DATE '2017-06-25' FROM DUAL UNION ALL
SELECT 'H', DATE '2017-06-24', DATE '2017-06-28' FROM DUAL UNION ALL -- Overlaps previous and next
SELECT 'I', DATE '2017-06-27', DATE '2017-06-30' FROM DUAL UNION ALL
SELECT 'J', DATE '2017-06-27', DATE '2017-06-28' FROM DUAL; -- Within H and I
Query:
SELECT SUM( days ) AS total_days
FROM (
SELECT dt - LAG( dt ) OVER ( ORDER BY dt ) + 1 AS days,
start_end
FROM (
SELECT dt,
CASE SUM( value ) OVER ( ORDER BY dt ASC, value DESC, ROWNUM ) * value
WHEN 1 THEN 'start'
WHEN 0 THEN 'end'
END AS start_end
FROM your_table
UNPIVOT ( dt FOR value IN ( start_date AS 1, end_date AS -1 ) )
)
WHERE start_end IS NOT NULL
)
WHERE start_end = 'end';
Output:
TOTAL_DAYS
----------
25
Explanation:
SELECT dt, value
FROM your_table
UNPIVOT ( dt FOR value IN ( start_date AS 1, end_date AS -1 ) )
This will UNPIVOT the table so that the start and end dates are in the same column (dt) and are given a corresponding value of +1 for a start and -1 for an end date.
SELECT dt,
SUM( value ) OVER ( ORDER BY dt ASC, value DESC, ROWNUM ) AS total,
value
FROM your_table
UNPIVOT ( dt FOR value IN ( start_date AS 1, end_date AS -1 ) )
Will give the start and end dates and the cumulative sum of those generated values. The start of a range will always have value=1 and total=1 and the end of a range will always have total=0. If a date is mid-way through a range then it will either have total>1 or value=-1 and total=1. Using this, if you multiply value and total then the start of a range is when value*total=1 and the end of a range is when value*total=0 and any other value indicates a date that is midway through a range.
Which is what this gives:
SELECT dt,
CASE SUM( value ) OVER ( ORDER BY dt ASC, value DESC, ROWNUM ) * value
WHEN 1 THEN 'start'
WHEN 0 THEN 'end'
END AS start_end
FROM your_table
UNPIVOT ( dt FOR value IN ( start_date AS 1, end_date AS -1 ) )
You can then filter out the dates when the start_end is NULL which will leave you with a table with alternating start and end rows which you can use LAG to calculate the number of days difference:
SELECT dt - LAG( dt ) OVER ( ORDER BY dt ) + 1 AS days,
start_end
FROM (
SELECT dt,
CASE SUM( value ) OVER ( ORDER BY dt ASC, value DESC, ROWNUM ) * value
WHEN 1 THEN 'start'
WHEN 0 THEN 'end'
END AS start_end
FROM your_table
UNPIVOT ( dt FOR value IN ( start_date AS 1, end_date AS -1 ) )
)
WHERE start_end IS NOT NULL
All you need to do then is to SUM all the differences for the end - start; which gives the query above.
As #Pravin Satav addressed, your requirement it's not very clear, something like this is what I understood from your explanation:
SELECT sum(CASE WHEN end_date=start_date THEN 1 ELSE (end_date-start_date)+1 END) as total_days
FROM my_table
WHERE <conditions that determine your "sample data">;
Need Suggestion to make it dynamic On Dates.
Expected:
Date, Total Sellers, Sellers From Previous Date
Currently:
Data in table(active_seller_codes): date, seller_code
Queries:
-- Date Wise Sellers Count
select date,count(distinct seller_code) as Sellers_COunt
from active_seller_codes where date between '2016-12-15' AND '2016-12-15'
-- Sellers from previous Days
select date,count(distinct seller_code) as Last_Day_Seller
from active_seller_codes
where date between '2016-12-15' AND '2016-12-15'
and seller_code IN(
select seller_code from active_seller_codes
where date between '2016-12-14' AND '2016-12-14'
)
group by 1
Database Using: Vertica
Reading attentively, you seem to want one row in the report, with the data from the search date in the first two columns and the data of the day before the search date in the third and fourth column, like so:
sales_date|sellers_count|prev_date |prev_sellers_count
2016-12-15| 8|2016-12-14| 5
The solution could be something like this (without the first Common Table Expression, which, in my case, contains the data, but in your case, the data would be in your active_seller_codes table.
WITH
-- initial input
(sales_date,seller_code) AS (
SELECT DATE '2016-12-15',42
UNION ALL SELECT DATE '2016-12-15',43
UNION ALL SELECT DATE '2016-12-15',44
UNION ALL SELECT DATE '2016-12-15',45
UNION ALL SELECT DATE '2016-12-15',46
UNION ALL SELECT DATE '2016-12-15',47
UNION ALL SELECT DATE '2016-12-15',48
UNION ALL SELECT DATE '2016-12-15',49
UNION ALL SELECT DATE '2016-12-14',42
UNION ALL SELECT DATE '2016-12-14',44
UNION ALL SELECT DATE '2016-12-14',46
UNION ALL SELECT DATE '2016-12-14',48
UNION ALL SELECT DATE '2016-12-14',50
UNION ALL SELECT DATE '2016-12-13',42
UNION ALL SELECT DATE '2016-12-13',43
UNION ALL SELECT DATE '2016-12-13',44
UNION ALL SELECT DATE '2016-12-13',45
UNION ALL SELECT DATE '2016-12-13',46
UNION ALL SELECT DATE '2016-12-13',47
UNION ALL SELECT DATE '2016-12-13',48
UNION ALL SELECT DATE '2016-12-13',49
)
,
-- search argument this, in the real query, would come just after the WITH keyword
-- as the above would be the source table
search_dt(search_dt) AS (SELECT DATE '2016-12-15')
,
-- the two days we're interested in, de-duped
distinct_two_days AS (
SELECT DISTINCT
sales_date
, seller_code
FROM active_seller_codes
WHERE sales_date IN (
SELECT search_dt FROM search_dt -- the search date
UNION ALL SELECT search_dt - 1 FROM search_dt -- the day before
)
)
,
-- the two days we want one above the other,
-- with index for the final pivot
vertical AS (
SELECT
ROW_NUMBER() OVER (ORDER BY sales_date DESC) AS idx
, sales_date
, count(DISTINCT seller_code) AS seller_count
FROM distinct_two_days
GROUP BY 2
)
SELECT
MAX(CASE idx WHEN 1 THEN sales_date END) AS sales_date
, SUM(CASE idx WHEN 1 THEN seller_count END) AS sellers_count
, MAX(CASE idx WHEN 2 THEN sales_date END) AS prev_date
, SUM(CASE idx WHEN 2 THEN seller_count END) AS prev_sellers_count
FROM vertical
;
sales_date|sellers_count|prev_date |prev_sellers_count
2016-12-15| 8|2016-12-14| 5
I have a table and each record has a date. We can assume that a date range is contiguous if there's not a 3 month break. How can I find the start of the most recent contiguous date range?
For example, imagine if I had this data:
1990-5-1
1990-6-4
1990-10-28
1990-11-14
1990-12-19
1991-1-20
1991-4-30
1991-5-13
I'd like for it to return 1991-4-30 because it's the start of the most recent contiguous range of dates.
I think this does what you're looking for. Using my own table and column names as test data. This is on Oracle.
select * from (
select * from sm_ss_tickets t1 where exists (
select * from sm_ss_tickets t2 where t2.created_date between t1.created_date and t1.created_date+90 and t1.rowid <> t2.rowid
) order by created_date asc
) where rownum = 1;
Maybe something like the following would work:
WITH d1 AS (
SELECT date'1990-05-01' AS dt FROM dual
UNION ALL
SELECT date'1990-06-04' AS dt FROM dual
UNION ALL
SELECT date'1990-10-28' AS dt FROM dual
UNION ALL
SELECT date'1990-11-14' AS dt FROM dual
UNION ALL
SELECT date'1990-12-19' AS dt FROM dual
UNION ALL
SELECT date'1991-01-20' AS dt FROM dual
UNION ALL
SELECT date'1991-04-30' AS dt FROM dual
UNION ALL
SELECT date'1991-05-13' AS dt FROM dual
)
SELECT MAX(dt) FROM (
SELECT dt, LAG(dt) OVER ( ORDER BY dt ) AS prev_dt, LEAD(dt) OVER ( ORDER BY dt ) AS next_dt
FROM d1
) WHERE ( dt > ADD_MONTHS(prev_dt, 3) OR prev_dt IS NULL )
AND dt > ADD_MONTHS(next_dt, -3)
In the above, a date can only be the start of a contiguous sequence if there is no prior date within 3 months (either it is more than three months ago or it doesn't exist at all) and there is also a subsequent date within 3 months.
You can use LAG and LEAD. Find the query below. I think it works fine.
tmp_year is the table I have created. tdate is the column.
The records in the table are
28-JAN-15
27-JAN-15
26-JAN-15
25-JAN-15
12-JUL-14
11-JUL-14
10-JUL-14
09-JUL-14
24-DEC-13
23-DEC-13
22-DEC-13
21-DEC-13
15-SEP-13
07-JUN-13
27-FEB-13
19-NOV-12
11-AUG-12
Please find the query which returns 25th Jan 2015.
select max(d.tdate) from (
select c.tdate,c.next_date,c.date_diff,lag(date_diff) over( order by tdate) prev_diff from (
select b.tdate ,b.next_date,(next_date-tdate) date_diff from
(select a.tdate,lead(a.tdate) over(order by a.tdate) next_date from tmp_year a ) b ) c) d where d.date_diff<90 and d.prev_diff>=90;