SQL：按比某物大多少排序？
我这里有两个 datetime 字段：actual_delivery 和 scheduled_delivery
我想要执行的 ORDER BY 是多少 actual_delivery code> 大于 scheduled_delivery。
我在本地使用 MySQL，在生产中使用 PostgreSQL，因此它需要同时适用于两者。
I have two datetime fields here: actual_delivery and scheduled_delivery
What I want to do an ORDER BY on is how much great actual_delivery is than scheduled_delivery.
I'm using MySQL locally and PostgreSQL in production, so it needs to work for both.
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如果我在 SQL Server 中执行此操作，我会计算 DATEDIFF(actual_delivery, Scheduled_deliver) AS [DeliveryDifference]，然后按该计算列进行排序。
快速搜索表明 MySql 中有一个
datediff函数，但 PostgreSQL 中的语法可能略有不同，因此您可能必须在那里创建自己的函数。If I were doing it in SQL Server I'd calculate
DATEDIFF(actual_delivery, scheduled_deliver) AS [DeliveryDifference]then order by that computed column.A quick search indicates there's a
datedifffunction in MySql but the syntax may be slightly different in PostgreSQL so you may have to create your own function there.试试这个：
SELECTactual_delivery、scheduled_delivery、actual_delivery-scheduled_delivery作为差异FROM表名ORDER BY差异Try this:
SELECT actual_delivery, scheduled_delivery, actual_delivery - scheduled_delivery as difference FROM tablename ORDER BY difference