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.
Related
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
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
)
Using
https://data.seattle.gov/Public-Safety/PDRs-After-using-City-of-Seattle-Public-Records-Re/wj44-r6br/data I want the know on each date the number of public disclosure requests were open. This means per date I want the number of requests created before or same day as date and don't have a close date after the date.
I copied it to https://data.world/timacbackup/seattle-police-public-disclosure-requests where I can use SQL.
The closest I've gotten is
SELECT CAST(seattle_police_records_requests.request_create_date AS DATE) AS the_date,
count(*)
FROM seattle_police_records_requests
GROUP BY CAST(seattle_police_records_requests.request_create_date AS DATE)
ORDER BY the_date DESC;
I tried
SELECT CAST(request_create_date AS DATE) AS the_date,
count((
SELECT request_create_date
FROM seattle_police_records_requests AS t
WHERE CAST(t.request_create_date AS DATE) < d.request_create_date
))
FROM seattle_police_records_requests AS d
GROUP BY CAST(request_create_date AS DATE)
ORDER BY the_date DESC;
but get unknown table 'd' for the count subquery.
The last query I tried is
WITH dates
AS (
SELECT CAST(request_create_date AS DATE) AS create_date,
CAST(request_closed_date AS DATE) AS closed_date
FROM seattle_police_records_requests
),
create_dates
AS (
SELECT DISTINCT CAST(request_create_date AS DATE) AS create_date
FROM seattle_police_records_requests
)
SELECT create_dates.create_date,
COUNT(*)
FROM dates
INNER JOIN create_dates ON dates.create_date = create_dates.create_date
GROUP BY create_dates.create_date
HAVING dates.create_date <= create_dates.create_date
ORDER BY create_dates.create_date DESC
and basically it's just counting # of requested opened on given day not all that were open as of given day.
After importing the "created" and "closed" values into SQL Server as datetime columns I was able to generate the counts per day like so:
WITH
given_dates AS
(
SELECT DISTINCT CAST(created AS DATE) AS given_date
FROM seattle_police_records_requests
)
SELECT
given_date,
(
SELECT COUNT(*)
FROM seattle_police_records_requests
WHERE created <= DATEADD(DAY, 1, given_date) AND (closed > given_date OR closed IS NULL)
) AS num_open
FROM given_dates
ORDER BY given_date;
The DATEADD was necessary to include requests opened during that day, since the comparison of a date and a datetime implies that the date value is midnight (i.e., the very beginning of that day).
How can I make a query in SQL Server to query for all rows for the next 5 days.
The problem is that it has to be days with records, so the next 5 days, might become something like, Today, Tomorrow, some day in next month, etc...
Basically I want to query the database for the records for the next non empty X days.
The table has a column called Date, which is what I want to filter.
Why not split the search into 2 queries. First one searches for the date part, the second uses that result to search for records IN the dates returned by the first query.
#Anagha is close, just a little modification and it is OK.
SELECT *
FROM TABLE
WHERE DATE IN (
SELECT DISTINCT TOP 5 DATE
FROM TABLE
WHERE DATE >= referenceDate
ORDER BY DATE
)
You can use following SQL query where 5 different dates are fetched at first then all rows for those selected dates are displayed
declare #n int = 5;
select *
from myData
where
datecol in (
SELECT distinct top (#n) cast(datecol as date) as datecol
FROM myData
WHERE datecol >= '20180101'
ORDER BY datecol
)
Try this:
select date from table where date in (select distinct top 5 date
from table where date >= getdate() order by date)
If your values are dates, you can use `dense_rank():
select t.*
from (select t.*, dense_rank() over (order by datecol) as seqnum
from t
where datecol >= cast(getdate() as date)
) t
where seqnum <= 5;
If the column has a time component and you still want to define days by midnight-to-midnight (as suggested by the question), just convert to date:
select t.*
from (select t.*,
dense_rank() over (order by cast(datetimecol as date)) as seqnum
from t
where datetimecol >= cast(getdate() as date)
) t
where seqnum <= 5;
I have a table, lets say "Records" with structure:
id date
-- ----
1 2012-08-30
2 2012-08-29
3 2012-07-25
I need to write an SQL query in PostgreSQL to get record_id for MIN date in each month.
month record_id
----- ---------
8 2
7 3
as we see 2012-08-29 < 2012-08-30 and it is 8 month, so we should show record_id = 2
I tried something like this,
SELECT
EXTRACT(MONTH FROM date) as month,
record_id,
MIN(date)
FROM Records
GROUP BY 1,2
but it shows 3 records.
Can anybody help?
SELECT DISTINCT ON (EXTRACT(MONTH FROM date))
id,
date
FROM Records1
ORDER BY EXTRACT(MONTH FROM date),date
SQLFiddle http://sqlfiddle.com/#!12/76ca2/3
UPD: This query:
1) Orders the records by month and date
2) For every month picks the first record (the first record has MIN(date) because of ordering)
Details here http://www.postgresql.org/docs/current/static/sql-select.html#SQL-DISTINCT
This will return multiples if you have duplicate minimum dates:
Select
minbymonth.Month,
r.record_id
From (
Select
Extract(Month From date) As Month,
Min(date) As Date
From
records
Group By
Extract(Month From date)
) minbymonth
Inner Join
records r
On minbymonth.date = r.date
Order By
1;
Or if you have CTEs
With MinByMonth As (
Select
Extract(Month From date) As Month,
Min(date) As Date
From
records
Group By
Extract(Month From date)
)
Select
m.Month,
r.record_id
From
MinByMonth m
Inner Join
Records r
On m.date = r.date
Order By
1;
http://sqlfiddle.com/#!1/2a054/3
select extract(month from date)
, record_id
, date
from
(
select
record_id
, date
, rank() over (partition by extract(month from date) order by date asc) r
from records
) x
where r=1
order by date
SQL Fiddle
select distinct on (date_trunc('month', date))
date_trunc('month', date) as month,
id,
date
from records
order by 1, 3 desc
I think you need use sub-query, something like this:
SELECT
EXTRACT(MONTH FROM r.date) as month,
r.record_id
FROM Records as r
INNER JOIN (
SELECT
EXTRACT(MONTH FROM date) as month,
MIN(date) as mindate
FROM Records
GROUP BY EXTRACT(MONTH FROM date)
) as sub on EXTRACT(MONTH FROM r.date) = sub.month and r.date = sub.mindate