WHERE datetime is 08:00 in MySQL-CodePudding

I have a following problem. I have this table:

day | valueA | valueB
2022-04-24 04:35:08 2929 | 5845
2022-04-24 06:30:10 2929 | 5844
2022-04-24 07:25:12 2929 | 5844
2022-04-24 08:00:12 2929 | 7844
2022-04-24 12:15:10 2929 | 5844
2022-04-24 14:10:09 2929 | 5844

I would like to get valueB - valueA only if it is 8:00. So from the example above I would get 4915, because 7844-2929.

I tried this based on this answer, which is however about MSSQL: Extracting hours from a DateTime (SQL Server 2005) but it does not work:

SELECT valueB - valueA AS diff
FROM table
WHERE day(HOUR, GETDATE()) = 8

I got that MySQLdb._exceptions.OperationalError: (1305, 'FUNCTION table.day does not exist'). How can I do this in MySQL, please?

CodePudding user response：

Try

SELECT valueB - valueA AS diff
FROM table
WHERE HOUR(day) = 8

CodePudding user response：

If you are only interested in 8:00 you also need to check for the minutes

SELECT valueB - valueA AS diff
FROM table
WHERE HOUR(day) = 8 AND MINUTE(day) = 0

CodePudding user response：

That should work:

SELECT valueB - valueA AS diff
FROM table
WHERE HOUR(day) = 8