我正在尝试在某些日期执行减法运算，在我的特定情况下，我只需要减去工作日（不包括周末，即周日和周六）
我的桌子是这样的：
date | days_to_subtract | dayName
23-04-2004 | 3 | Friday
23-05-2004 | 5 | Monday
21-04-2004 | 7 | Tuesday
25-04-2004 | 30 | Monday
01-04-2004 | 22 | Thursday
我的目标是计算一个新的列（新日期），从日期减去天数，只考虑工作（劳动）日，不包括（或忽略）周末日。
我试过这样的方法：
SELECT
CASE WHEN dayName = "Monday" then date_sub(date, days_to_subtract + 2)
WHEN dayName = "Tuesday" then date_sub(date, days_to_subtract + 1)
ELSE date_sub(date, days_to_subtract)
END AS newColumn
FROM
"MyTable"
但当减天数大于5时，这将不起作用。
有没有建议使用简单的SQL和不复杂的函数来解决这个问题？
最佳答案
我们先来解释一下你的情况：
星期一）Suposse我们在星期一定位，那么当减去N（虚拟）天时，您期望（实际）减去的结果如下：
"Mon" - 1 day => "Mon" - (1 + 2 * (1)) days => "Mon" - 3 days /* 1 weekend affecting */
"Mon" - 2 days => "Mon" - (2 + 2 * (1)) days => "Mon" - 4 days /* 1 weekend affecting */
"Mon" - 3 days => "Mon" - (3 + 2 * (1)) days => "Mon" - 5 days /* 1 weekend affecting */
...
"Mon" - 6 days => "Mon" - (6 + 2 * (2)) days => "Mon" - 10 days /* 2 weekends affecting */
...
星期二）Suposse我们定位在星期二，那么当减去N（虚拟）天时，您期望（实际）减去的结果如下：
"Tue" - 1 day => "Tue" - (1 + 2 * (0)) days => "Tue" - 1 days /* 0 weekends affecting */
"Tue" - 2 days => "Tue" - (2 + 2 * (1)) days => "Tue" - 4 days /* 1 weekend affecting */
"Tue" - 3 days => "Tue" - (3 + 2 * (1)) days => "Tue" - 5 days /* 1 weekend affecting */
...
"Tue" - 6 days => "Tue" - (6 + 2 * (1)) days => "Tue" - 8 days /* 1 weekend affecting */
"Tue" - 7 days => "Tue" - (7 + 2 * (2)) days => "Tue" - 11 days /* 2 weekends affecting */
...
星期三）Suposse我们定位在星期三，那么当减去N（虚拟）天时，您希望（实际）减去的结果如下：
"Wed" - 1 day => "Wed" - (1 + 2 * (0)) days => "Wed" - 1 days /* 0 weekends affecting */
"Wed" - 2 days => "Wed" - (2 + 2 * (0)) days => "Wed" - 2 days /* 0 weekends affecting */
"Wed" - 3 days => "Wed" - (3 + 2 * (1)) days => "Wed" - 5 days /* 1 weekend affecting */
...
"Wed" - 6 days => "Wed" - (6 + 2 * (1)) days => "Wed" - 7 days /* 1 weekend affecting */
"Wed" - 7 days => "Wed" - (7 + 2 * (1)) days => "Wed" - 9 days /* 1 weekend affecting */
"Wed" - 8 days => "Wed" - (8 + 2 * (2)) days => "Wed" - 12 days /* 2 weekends affecting */
...
我们可以注意到，这个逻辑中间有一个数学演算。如果我们以这种方式索引一周中的几天：
0 <-> Monday
1 <-> Tuesday
2 <-> Wednesday
3 <-> Thursday
4 <-> Friday
那么，一个与之前的微积分相匹配的公式是：
Week_Day - X days => Week_Day - X days + 2 * CEIL((X - INDEX_OF(Week_Day)) / 5) days
where INDEX_OF() is a function that returns the previous indexing for days.
例1）
假设我们需要从星期三减去两天，那么前面的公式将缩减为：
"Wed" - 2 days => "Wed" - 2 + 2 * CEIL((2 - 2) / 5)
"Wed" - 2 days => "Wed" - 2 + 2 * CEIL(0 / 5)
"Wed" - 2 days => "Wed" - 2 + 2 * 0
"Wed" - 2 days => "Wed" - 2 + 0
"Wed" - 2 days => "Wed" - 2 days
例2）
假设我们需要从星期三减去8天（在这种情况下，我们有两个周末在减法的中间），那么前面的公式将减少为：
"Wed" - 8 days => "Wed" - 8 + 2 * CEIL((8 - 2) / 5)
"Wed" - 8 days => "Wed" - 8 + 2 * CEIL(6 / 5)
"Wed" - 8 days => "Wed" - 8 + 2 * 2 /* Two weekends are affecting */
"Wed" - 8 days => "Wed" - 8 + 4
"Wed" - 8 days => "Wed" - 12 days
因此，基于这个公式，并操作MySQL DAYOFWEEK（）方法来生成INDEX_OF（）函数，我们可以执行下一个查询：
SELECT
date AS OriginalDate,
DAYOFWEEK(date) - 2 AS IndexOfDay,
days_to_subtract AS DaysToSubtract,
2 * CEIL((days_to_subtract - (DAYOFWEEK(date) - 2)) / 5) AS WeekendDaysToAdd,
DATE_SUB(
date,
INTERVAL days_to_subtract + (2 * CEIL((days_to_subtract - (DAYOFWEEK(date) - 2)) / 5)) DAY
) AS CalucaltedDay
FROM
myTable
最后一列calculateDay将包含您希望从减法中获得的日期，其他列仅用于检查数学演算的步骤。我希望你能理解这背后的逻辑，因为我花了一些时间去弄清楚并试图解释它。您可以看到下一个链接的工作示例：
1）DB Fiddle
2）SQL Fiddle
关于mysql - 从忽略周末的日期开始减去几天，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/53015876/