我们先来解释一下你的情况：
星期一）假设我们的位置是星期一，那么当减去n（虚拟）天时，您期望（真实）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）假设我们的位置是星期二，那么当减去n（虚拟）天时，您期望的（真实）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）假设我们的位置是星期三，那么当减去n（虚拟）天时，您期望（真实）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，在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，与前面的演算相匹配的公式是：

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2天，那么前面的公式将简化为：

"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，我们在减去期间有两个周末），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（）方法来生成我们的index\u of（）函数，我们可以进行下一个查询：

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calculatedday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解释它。您可以在下面的链接中看到工作示例：
1） 小提琴
2） sql小提琴