MySQL - 使用 SET 语句的 UPDATE 查询取决于前一个 SET 语句的结果

MySQL - UPDATE query with SET statement dependent on the outcome of the previous SET statement(MySQL - 使用 SET 语句的 UPDATE 查询取决于前一个 SET 语句的结果)

问题描述

这是我希望通过 UPDATE 语句实现的目标的表格表示形式.

Here is a tabular representation of what I would like to achieve with an UPDATE statement.

+----+----+---+---+----+----------+---------------+---------------+
| ID | A  | B | C | D  |  Calc A  |    Calc B     |    Calc C     |
+----+----+---+---+----+----------+---------------+---------------+
|  1 |  6 | 5 | 2 | 10 | =[A]-[B] | =[Calc A]/[D] | =[B]/[Calc B] |
|  2 |  8 | 5 | 2 | 10 | =[A]-[B] | =[Calc A]/[D] | =[B]/[Calc B] |
|  3 | 10 | 5 | 2 | 10 | =[A]-[B] | =[Calc A]/[D] | =[B]/[Calc B] |
+----+----+---+---+----+----------+---------------+---------------+

实现此目的的当前 UPDATE 语句如下...

My Current UPDATE statement to achieve this is as follows...

UPDATE [EXAMPLE]
SET [Calc A]    = A - B
    , [Calc B]  = [Calc A] / D
    , [Calc C]  = B / [Calc B] 

但是它没有按预期工作.[Calc A] 将在第一次更新时正确计算.但是 [Calc B] 将使用 [Calc A] 中的 OLD 值进行计算，而不是我刚刚写入数据库的 NEW 更新值.这适用于 [Calc C]，它再次引用 [Calc B] 的 OLD 值.

However it is not working as intended. [Calc A] will calculate correctly on the first UPDATE. However [Calc B] will calculate using the OLD value in [Calc A] and not the NEW updated value I just wrote to the database. This holds true for [Calc C] which again refers to the OLD value of [Calc B].

如果您执行 UPDATE 语句 3 次，数据将正确计算出来.[Calc A] 在第一次计算中设置正确，然后 [Calc B] 将在第二次 UPDATE 中引用 [Calc A] 的正确更新值，然后 [Calc C] 将在 [Calc B] 中引用正确的值第三次更新.

If you perform the UPDATE statement 3 times the data will calculate out correctly. [Calc A] is set correctly in the first calculation, then [Calc B] will reference the correct updated value of [Calc A] in the second UPDATE, then [Calc C] will reference the correct value of [Calc B] in the 3rd UPDATE.

所以我的问题是如何在一个更新语句中将所有列设置为正确的值?

So my question is how do I set all the columns to their correct value in ONE update statement?

推荐答案

独立计算即可:

update [EXAMPLE]
set [Calc A] = A - B,
    [Calc B] = (A - B) / D,
    [Calc C] = B / ((A - B) / D)

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