WHERE 日期时间是 08:00 在 MySQL

Question

我有以下问题。我有这个 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

只有8:00我才想得到valueB - valueA。所以从上面的例子我会得到 4915，因为 7844-2929.

我根据这个答案尝试了这个，但是它是关于 MSSQL 的：Extracting hours from a DateTime (SQL Server 2005) 但它不起作用：

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

我明白了 MySQLdb._exceptions.OperationalError: (1305, 'FUNCTION table.day does not exist')。 MySQL 我该如何做？

Answer 1

尝试

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

Answer 2

如果您只对 8:00 感兴趣，您还需要查看会议记录

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

Answer 3

应该可行：

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

WHERE 日期时间是 08:00 在 MySQL

WHERE datetime is 08:00 in MySQL

mysql