最后一个工作日DB

Question

我这一周的所有天的数据，我想找到最近的工作日数据即dayofweek != 1 and dayofweek != 7

我笨拙的WHERE与CASE是沿着

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 

东西线

Answer 1

此代码，day(time) = date_sub(day(time), interval 2 day)从未匹配，因为day(time)从未有过等于day(time) - 1 day。这与在代数中说x=x-1一样......它不是真的。我怀疑你真正的意思是做这样的事情：

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. 

的date()功能消除了一部分时间，所以你可以就在一天进行比较。 The date_sub()函数需要从curdate()中减去以获取最近的工作日。在你的代码中，它是从存储在表中的DATETIME中减去的。