Question

我有一周中所有日子的数据，我想查找最近一个工作日的数据，dayofweek != 1 and dayofweek != 7

我的笨拙与CASE的关系是

WHERE CASE WHEN dayofweek(curdate()) = 1 THEN day(time) = date_sub(day(time), interval 2 day) 
    WHEN dayofweek(curdate()) = 7 THEN day(time) = date_sub(day(time), interval 1 day)  
    WHEN dayofweek(curdate()) != 7 AND dayofweek(curdate()) != 1 THEN day(time) = day(curdate())  ELSE 1 = 1  END

Answer 1

此代码day(time) = date_sub(day(time), interval 2 day)永远不会匹配，因为day(time) 永远不会 等于day(time) - 1 day。这与在代数中说x=x-1相同......它根本不可能是真的。我怀疑你实际上是想做这样的事情：

WHERE CASE WHEN dayofweek(curdate()) = 1 THEN date(time) = date(date_sub(curdate(), interval 2 day)) 
...
...
... This part was left as an exercise for the asker.

date()功能会删除时间部分，因此您只需比较一天。 The date_sub()函数需要从curdate()中减去才能获得最新的工作日。在您的代码中，它从表中存储的DATETIME中减去。

DB的最后一个工作日

1 个答案: