'Group by month taking into account on timezone (SQL Server)
I've seen the following question about grouping by month:
DATEADD(MONTH, DATEDIFF(MONTH, 0, Closing_Date), 0)
How to group by month from Date field using sql
This works - but I want to run this against a specific timezone, taking DST into account.
To be specific, the Netherlands timezone: Central European Standard Time (GMT+1).
This will be GMT+2 sometimes with DST.
In my case I have an Orders table with an OrderDatetime which is a DateTimeOffset(7). all values are UTC (+00:00).
SELECT
DATEADD(MONTH, DATEDIFF(MONTH, 0, [Orders].[OrderDateTime]), 0) AS [Month],
...
FROM
[Orders]
GROUP BY
DATEADD(MONTH, DATEDIFF(MONTH, 0, [Orders].[OrderDateTime]), 0)
What would such query look like when taking into account a specific timezone?
Bonus: I'd preferably make this configurable, so the timezone would be a parameter - what would such query look like?
Solution 1:[1]
You can use SWITCHOFFSET
CREATE TABLE test
(
OrderDateTime datetimeoffset
, tz varchar (8)
);
INSERT INTO test
VALUES ('2022-01-31 17:00:00 -5:00', '+08:00');
SELECT
DATEADD(MONTH, DATEDIFF(MONTH, 0, [OrderDateTime]), 0) AS [Month1]
, DATEADD(MONTH, DATEDIFF(MONTH, 0, SWITCHOFFSET (OrderDateTime, tz)), 0) AS [Month2]
FROM test;
Returns
Month1 Month2
2022-01-01 00:00:00.000 2022-02-01 00:00:00.000
Solution 2:[2]
If the value is already a datetimeoffset then SQL Server already takes timezones into account, so I suspect the problem you think exists, doesn't.
Take the following:
CREATE TABLE dbo.YourTable (YourColumn datetimeoffset(7));
INSERT INTO dbo.YourTable (YourColumn)
VALUES('2022-01-24T13:00:00+00:00'),
('2022-01-24T14:00:00+00:00'),
('2022-01-24T12:00:00+00:00');
GO
SELECT *
FROM dbo.YourTable
WHERE YourColumn = CONVERT(datetimeoffset(7),'2022-01-24T14:00:00+01:00');
GO
DROP TABLE dbo.YourTable;
GO
This returns the row with the value 2022-01-24T13:00:00.0000000 +00:00 because 2022-01-24T13:00:00.0000000 +00:00 and 2022-01-24T14:00:00+01:00 are the same time.
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 | Serg |
| Solution 2 | Larnu |
