我有以下問題。我有這張桌子：
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
我只想得到valueB - valueA它8:00。所以從上面的例子我會得到 4915，因為 7844-2929。
我根據這個答案嘗試了這個，但是關于 MSSQL: 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')。請問如何在 MySQL 中做到這一點？
uj5u.com熱心網友回復：
嘗試
SELECT valueB - valueA AS diff
FROM table
WHERE HOUR(day) = 8
uj5u.com熱心網友回復：
如果您只對 8:00 感興趣，您還需要查看分鐘
SELECT valueB - valueA AS diff
FROM table
WHERE HOUR(day) = 8 AND MINUTE(day) = 0
uj5u.com熱心網友回復：
那應該作業：
SELECT valueB - valueA AS diff
FROM table
WHERE HOUR(day) = 8
轉載請註明出處，本文鏈接：https://www.uj5u.com/qiye/464322.html
標籤：mysql