I have the following table in sql server database environment.
the format of start date MM/DD/YYYY.
I need the result to be like the following table.
based on start date column the record should segregated to each month in the period between start date and end date
You can use a recursive CTE:
with cte as (
select id, startdate as dte, enddate
from t
union all
select id,
dateadd(day, 1, eomonth(dte)),
enddate
from t
where eomonth(dte) < enddate
)
select id, dte,
lead(dte, 1, enddate) over (partition by id order by dte)
from cte;
Thank you Gordon Linoff
Using CTE I have got the following result
My code
WITH cte
AS (SELECT 1 AS id,
Cast('2010-01-20' AS DATE) AS trg,
Cast('2010-01-20' AS DATE) AS strt_dte,
Cast('2010-03-15' AS DATE) AS end_dte
UNION ALL
SELECT id,
Dateadd(day, 1, Eomonth (trg)),
strt_dte,
end_dte
FROM cte
WHERE Eomonth(trg) < end_dte)
SELECT id,
trg,
strt_dte,
end_dte,
Lead (trg, 1, end_dte)
OVER (
partition BY id
ORDER BY trg) AS lead_result
FROM cte
Related
I have a table that where each record has start and end dates. I need to return a record for each day between start and end date fields (including the start and end dates). Using MS SQL Server.
Example:
Current data
Data required:
Looking for recommendations.
Thanks.
You can use recursive cte :
with cte as (
select id, startdate, enddate, startdate as date
from table t
union all
select id, startdate, enddate, dateadd(day, 1, date)
from cte c
where date < enddate
)
select *
from cte c
option (maxrecursion 0);
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
Having Lead function and doing calculation(deduct) in ONE select statement instead of using nested select statement to achieve the results
Below shows my current not working sql statement. CurrentDate - Nxcurrentdate and having a column with showing the difference:
SELECT
ID,
CURRENTDATE,
LEAD(CURRENTDATE,1) OVER ( PARTITION BY ID ORDER BY ID, CURRENTDATE ) NX_DATE,
NX_DATE - CURRENTDATE AS DATE_DIFF
FROM TABLEA
Expected results should be:
ID CurrentDate NxDate DateDiff
you can use datediff(interval, date1, date2) and getdate()
Select *, DATEDIFF(day, t1.CURRENTDATE, t1.NX_DATE) AS DATE_DIFF) from (
SELECT
ID, getdate() as CURRENTDATE,
LEAD(getdate(),1) OVER ( PARTITION BY ID ORDER BY ID, getdate()) NX_DATE
FROM TABLEA) t1
You can try below:
SELECT
ID, CURRENTDATE,
LEAD(CURRENTDATE,1) OVER ( PARTITION BY ID ORDER BY ID, CURRENTDATE ) NX_DATE,
datediff(day,CURRENTDATE,LEAD(CURRENTDATE,1) OVER ( PARTITION BY ID ORDER BY ID, CURRENTDATE)) AS DATE_DIFF
from tablename
For example
If I want to check in every day last week
select count(ID) from DB where date < "2019/07/01"
select count(ID) from DB where date < "2019/07/02"
select count(ID) from DB where date < "2019/07/03"
...
select count(ID) from DB where date < "2019/07/08"
like
0701 10
0702 15
0703 23
...
0707 45
How to do this without loop and one query?
You can generate the dates using a recursive CTE (or other method) and then run the query:
with dates as (
select convert(date, '2019-07-01') as dte union all
select dateadd(day, 1, dte)
from dates
where dte < '2019-07-08'
)
select d.dte,
(select count(*) from DB where DB.date < d.dte)
from dates d;
More efficient, though, is a cumulative sum:
select db.*
from (select date, count(*) as cnt, sum(count(*)) over (order by date) as running_cnt
from db
group by date
) d
where d.date >= '2019-07-01' and d.date < '2019-07-09';
Are you just counting the number by day?
Something like
SELECT MONTH(date), DAY(date), COUNT(ID)
FROM DB
GROUP BY MONTH(date), DAY(date);
(assuming date is a DATE or DATETIME)
Do it with window Count. range between current row and current row selects exactly this day rows.
select distinct date, count(1) over (order by Date) - count(1) over (order by Date range between current row and current row)
from DB
where date between '2019-07-01' and '2019-07-08';
I assume date column is exactly DATE.
I have a table named Employees with Columns: PersonID, Name, StartDate. I want to calculate 1) difference in days between the newest and oldest employee and 2) the longest period of time (in days) without any new hires. I have tried to use DATEDIFF, however the dates are in a single column and I'm not sure what other method I should use. Any help would be greatly appreciated
Below is for BigQuery Standard SQL
#standardSQL
SELECT
SUM(days_before_next_hire) AS days_between_newest_and_oldest_employee,
MAX(days_before_next_hire) - 1 AS longest_period_without_new_hire
FROM (
SELECT
DATE_DIFF(
StartDate,
LAG(StartDate) OVER(ORDER BY StartDate),
DAY
) days_before_next_hire
FROM `project.dataset.your_table`
)
You can test, play with above using dummy data as in the example below
#standardSQL
WITH `project.dataset.your_table` AS (
SELECT DATE '2019-01-01' StartDate UNION ALL
SELECT '2019-01-03' StartDate UNION ALL
SELECT '2019-01-13' StartDate
)
SELECT
SUM(days_before_next_hire) AS days_between_newest_and_oldest_employee,
MAX(days_before_next_hire) - 1 AS longest_period_without_new_hire
FROM (
SELECT
DATE_DIFF(
StartDate,
LAG(StartDate) OVER(ORDER BY StartDate),
DAY
) days_before_next_hire
FROM `project.dataset.your_table`
)
with result
Row days_between_newest_and_oldest_employee longest_period_without_new_hire
1 12 9
Note use of -1 in calculating longest_period_without_new_hire - it is really up to you to use this adjustment or not depends on your preferences of counting gaps
1) difference in days between the newest and oldest record
WITH table AS (
SELECT DATE(created_at) date, *
FROM `githubarchive.day.201901*`
WHERE _table_suffix<'2'
AND repo.name = 'google/bazel-common'
AND type='ForkEvent'
)
SELECT DATE_DIFF(MAX(date), MIN(date), DAY) max_minus_min
FROM table
2) the longest period of time (in days) without any new records
WITH table AS (
SELECT DATE(created_at) date, *
FROM `githubarchive.day.201901*`
WHERE _table_suffix<'2'
AND repo.name = 'google/bazel-common'
AND type='ForkEvent'
)
SELECT MAX(diff) max_diff
FROM (
SELECT DATE_DIFF(date, LAG(date) OVER(ORDER BY date), DAY) diff
FROM table
)