Select Table of Days For One Year SQLITE - sql

I am trying to select a table of every date for the last year.
In SQL Server, I can run something like this:
SELECT TOP (DATEDIFF(DAY, DATEADD(YEAR, - 1, GETDATE()), GETDATE()) + 1)
Date = CAST(DATEADD(DAY, ROW_NUMBER() OVER (ORDER BY a.object_id) - 1, DATEADD(YEAR, - 1, GETDATE())) AS DATE)
FROM sys.all_objects a
It returns 1 column with 366 rows containing the dates from 1 year ago until now.
I am looking for something equivalent in SQLITE.

You can do it with a recursive CTE:
WITH cte AS (
SELECT DATE(CURRENT_DATE, '-1 year') date
UNION ALL
SELECT DATE(date, '+1 day')
FROM cte
WHERE date < CURRENT_DATE
)
SELECT * FROM cte;
See the demo.

You can also do this using an inline tally table
WITH L1(n) AS (
VALUES (1),(1),(1),(1),(1),(1),(1),(1),(1),(1),(1),(1),(1),(1),(1),(1),(1),(1),(1),(1)
),
L2(n) AS (
SELECT 1 FROM L1 a CROSS JOIN L1 b
)
SELECT DATE(CURRENT_DATE, (-ROW_NUMBER() OVER (ORDER BY 1) + 1) || ' days') AS date
FROM L2
LIMIT julianday(CURRENT_DATE) - julianday(DATE(CURRENT_DATE, '-1 year')) + 1;
db<>fiddle
For newer versions of SQLite you may want to add NOT MATERIALIZED to the CTEs.

Related

SQL Query to calculate the hours between two dates grouped by days

I need to write an SQL query for the following scenario.
I am having start date as 2020-01-10 13:00:00.347 and end date as 2020-01-12 02:00:00.347, so I need data grouped as
Day Hours
---- -----
10-01-2020 11
11-01-2020 24
12-01-2020 2.30
which means 11 hours was for the first date and 24 hours in second day and 2.3 hours on 3rd day.
What will the most Efficient SQL query to fetch the data in the above-mentioned format? Thanks in advance.
You can use a recursive CTE to break the dates into ranges:
with recursive cte as (
select start_date as day_start,
(case when date(start_date) = date(end_date) then end_date else date(start_date) + interval 1 day end) as day_end,
end_date
from (select cast('2020-01-10 13:00:00.347' as datetime) as start_date,
cast('2020-01-12 02:00:00.347' as datetime) as end_date
) t
union all
select day_end,
(case when date(day_end) = date(end_date) then end_date else date(day_end) + interval 1 day end) as day_end,
end_date
from cte
where day_end <> end_date
)
select day_start, day_end,
timestampdiff(second, day_start, day_end) / (60 * 60)
from cte;
Here is a db<>fiddle.
EDIT:
In SQL Server, this looks like:
with cte as (
select start_date as day_start,
(case when cast(start_date as date) = cast(end_date as date) then end_date else dateadd(day, 1, cast(start_date as date)) end) as day_end,
end_date
from (select cast('2020-01-10 13:00:00.347' as datetime) as start_date,
cast('2020-01-12 02:00:00.347' as datetime) as end_date
) t
union all
select day_end,
(case when cast(day_end as date) = cast(end_date as date) then end_date else dateadd(day, 1, day_end) end) as day_end,
end_date
from cte
where day_end <> end_date
)
select day_start, day_end,
datediff(second, day_start, day_end) / (60.0 * 60)
from cte;
Here is this db<>fiddle.
As the OP has asked for the most effecient method, and rCTE's are known to perform poorly, a more efficient approach would be using a Tally.
This isn't anywhere near as easy to read for a beginner, however, does get the results you are after (with the exception of that 2020-01-12 has a value of 2.0 not 2.3, as your math is clearly wrong there):
CREATE TABLE dbo.YourTable (StartDate datetime,
EndDate datetime);
INSERT INTO dbo.YourTable (StartDate,
EndDate)
VALUES('2020-01-10T13:00:00.347','2020-01-12T02:00:00.347'),
('2020-01-14T17:24:41.243','2020-01-19T09:17:12.997');
GO
WITH N AS(
SELECT N
FROM (VALUES(NULL),(NULL),(NULL),(NULL),(NULL),(NULL),(NULL),(NULL),(NULL),(NULL))N(N)),
Tally AS(
SELECT TOP(SELECT MAX(DATEDIFF(DAY, StartDate, EndDate)+1) FROM dbo.YourTable)
ROW_NUMBER() OVER (ORDER BY (SELECT NULL)) -1 AS I
FROM N N1, N N2, N N3), --1000 days enough?
Dates AS(
SELECT DATEADD(DAY, T.I,CONVERT(date,YT.StartDate)) AS [Date],
CASE WHEN T.I = 0 THEN YT.StartDate ELSE DATEADD(DAY, T.I,CONVERT(date,YT.StartDate)) END AS StartingDateTime,
CASE WHEN LEAD(T.I) OVER (PARTITION BY YT.StartDate ORDER BY T.I) IS NULL THEN YT.EndDate ELSE DATEADD(DAY, T.I+1,CONVERT(date,YT.StartDate)) END AS EndingDateTime
FROM Tally T
JOIN dbo.YourTable YT ON T.I <= DATEDIFF(DAY, YT.StartDate, YT.EndDate))
SELECT D.[Date],
(DATEDIFF(SECOND,D.StartingDateTime,D.EndingDateTime) * 1.0) / 60 / 60 AS [Hours]
FROM Dates D;
GO
DROP TABLE dbo.YourTable;
DB<>Fiddle

Redshift: Running query using GETDATE() at specified list of times

So, I have a query that uses GETDATE() in WHERE and HAVING clauses:
SELECT GETDATE(), COUNT(*) FROM (
SELECT 1 FROM events
WHERE (event_time > (GETDATE() - interval '25 hours'))
GROUP BY id
HAVING MAX(event_time) BETWEEN (GETDATE() - interval '25 hours') AND (GETDATE() - interval '24 hours')
)
I'm basically trying to find the number of unique ids that have their latest event_time between 25 and 24 hours ago with respect to the current time.
The problem: I have another table query_dts which contains one column containing timestamps. Instead of running the above query on the current time, using GETDATE(), I need to run in on the timestamp of every entry of the query_dts table. Any ideas?
Note: I'm not really storing query_dts anywhere. I've created it like this:
WITH query_dts AS (
SELECT (
DATEADD(hour,-(row_number() over (order by true)), getdate())
) as n
FROM events LIMIT 48
),
which I got from here
How about avoiding the generator altogether and instead just splitting the intervals:
SELECT
dateadd(hour, -distance, getdate()),
count(0) AS event_count
FROM (
SELECT
id,
datediff(hour, max(event_time), getdate()) AS distance
FROM events
WHERE event_time > getdate() - INTERVAL '2 days'
GROUP BY id) AS events_with_distance
GROUP BY distance;
You can use a JOIN to combine the two queries. Then you just need to substitute the values for your date expression. I think this is the logic:
WITH query_dts AS (
SELECT DATEADD(hour, -(row_number() over (order by true)), getdate()) as n
FROM events
LIMIT 48
)
SELECT d.n, COUNT(*)
FROM (SELECT d.n
FROM events e JOIN
query_dts d
WHERE e.event_time > d.n
GROUP BY id
HAVING MAX(event_time) BETWEEN n - interval '25 hours' AND n
) i;
Here's what I ended up doing:
WITH max_time_table AS
(
SELECT id, max(event_time) AS max_time
FROM events
WHERE (event_time > GETDATE() - interval '74 hours')
GROUP BY id
),
query_dts AS
(
SELECT (DATEADD(hour,-(row_number() over (ORDER BY TRUE) - 1), getdate()) ) AS n
FROM events LIMIT 48
)
SELECT query_dts.n, COUNT(*)
FROM max_time_table JOIN query_dts
ON max_time_table.max_time BETWEEN (query_dts.n - interval '25 hours') AND (query_dts.n - interval '24 hours')
GROUP BY query_dts.n
ORDER BY query_dts.n DESC
Here, I selected 74 hours because I wanted 48 hours ago + 25 hours ago = 73 hours ago.
The problem is that this isn't a general-purpose way of doing this. It's a highly specific solution for this particular problem. Can someone think of a more general way of running a query dependent on GETDATE() using a column of dates in another table?

Calculate Month Difference in Google BigQuery

BigQuery does not allow you to select the daypart MONTH in DATEDIFF.
Is there any other way to do this in BigQuery.
I can get days difference using the following but not months.
SELECT
OrderID,
OrderDate,
STRING( YEAR ( OrderDate )) + '-' + STRING(MONTH ( OrderDate )) as order_month,
UserID,
FirstOrderDate
DATEDIFF( OrderDate, FirstOrderDate) as date_diff,
FROM [orders.orders]
WHERE FirstOrderDate > DATE_ADD(CURRENT_TIMESTAMP(), -1, 'YEAR')
With Standard SQL you can use date_diff:
#StandardSQL
select date_diff(current_date, date '2018-03-06', month)
With standard SQL date_diff could work, but if you get an error saying "DATE_DIFF does not support the MONTH date part at [n:m]code", that's because your it's timestamp rather than date. To solve this issue, first cast the timestamp to date, then use date_diff function.
For example:
DATE_DIFF(CAST(OrderDate AS DATE), CAST(FirstOrderDate AS DATE), month) AS DATE_DIF
How about this?
(12 * YEAR(t2) + MONTH(t2)) - (12 * YEAR(t1) + MONTH(t1))
+ IF (DAY(t2) >= DAY(t1), 0, -1)
Running it over a couple of examples, it looks like what you'd want:
SELECT
(12 * YEAR(t2) + MONTH(t2)) - (12 * YEAR(t1) + MONTH(t1))
+ IF (DAY(t2) >= DAY(t1), 0, -1)
FROM
(SELECT TIMESTAMP("2010-12-01") as t1, TIMESTAMP("2011-01-01") as t2), // 1
(SELECT TIMESTAMP("2010-12-02") as t1, TIMESTAMP("2011-01-01") as t2), // 0
(SELECT TIMESTAMP("2010-12-01") as t1, TIMESTAMP("2011-12-01") as t2), // 12
(SELECT TIMESTAMP("2010-10-21") as t1, TIMESTAMP("2015-01-10") as t2) // 50

How to get the date for two saturdays ago

I have the following query which displays a table with date:
SELECT *
FROM [Db].[dbo].[btotals]
ORDER BY [Date] DESC
Which displays:
Date
06/07/2014
05/31/2014
05/24/2014
05/17/2014
05/10/2014
05/03/2014
If I pick SELECT TOP 1 will give me the first row. How can I modify my query so I get the week prior to last week? In this case the 5/31/14 row?
If your dates are always a week apart, and you just want the second row you can use ROW_NUMBER():
SELECT Date
FROM ( SELECT Date,
RowNumber = ROW_NUMBER() OVER(ORDER BY Date DESC)
FROM [Db].[dbo].[btotals]
) AS d
WHERE d.RowNumber = 2;
Otherwise you can use the following to get the saturday of 2 weeks ago:
SELECT DATEADD(DAY, -((DATEPART(WEEKDAY, GETDATE()) + ##DATEFIRST) % 7) - 7, CAST(GETDATE() AS DATE));
Then select your first date that is on or after that:
SELECT TOP 1 Date
FROM [Db].[dbo].[btotals]
WHERE Date >= DATEADD(DAY, -((DATEPART(WEEKDAY, GETDATE()) + ##DATEFIRST) % 7) - 7, CAST(GETDATE() AS DATE))
ORDER BY Date;
This should also work, if you are trying to select the second date, Though Gareth's approach of using ROW_NUNMBER is a better one.
SELECT TOP 1 *
FROM (
SELECT TOP 2 *
FROM [Db].[dbo].[btotals]
ORDER BY [Date] DESC
) as X
ORDER BY Date ASC
Another approach:
SELECT TOP 1 *
FROM [Db].[dbo].[btotals]
WHERE [Date] < (SELECT MAX([Date]) FROM [Db].[dbo].[btotals])
ORDER BY [Date] DESC

Get a table of dates of last 14 days (2 weeks) starting from previous day

I want to write a query which gives a coloumn of DATES of last 14 days
starting from yesterday.
Example:
Dates
2012-06-21
2012-06-20
2012-06-19
--
-
;WITH n(n) AS
(
SELECT TOP (14) ROW_NUMBER() OVER (ORDER BY [object_id])
FROM sys.objects ORDER BY [object_id]
)
SELECT Dates = DATEADD(DAY, -n, DATEDIFF(DAY, 0, GETDATE())) FROM n
ORDER BY n;
Another way;
;with days(day) as
(
select getdate() - 1 as day
union all
select day - 1
from days
where day > dateadd(day, -14, getdate())
)
select cast(day as date) from days