我需要计算MySQL中两个日期之间的差异(以天为单位),不包括周末(周六和周日).也就是说,天数之差减去周六和周日之间的数量.

目前,我只计算使用日期：

SELECT DATEDIFF('2012-03-18', '2012-03-01')

这个返回17,但我想排除周末,所以我想要12(因为第3和第4,第10和第11和第17是周末的日子).

我不知道从哪里开始.我知道WEEKDAY()函数和所有相关函数,但我不知道如何在这种情况下使用它们.

解决方法:

插图：

mtwtfSSmtwtfSS
  123456712345   one week plus 5 days, you can remove whole weeks safely
  12345-------   you can analyze partial week's days at start date
  -------12345   or at ( end date - partial days )

伪代码：

@S          = start date
@E          = end date, not inclusive
@full_weeks = floor( ( @E-@S ) / 7)
@days       = (@E-@S) - @full_weeks*7   OR (@E-@S) % 7

SELECT
  @full_weeks*5 -- not saturday+sunday
 +IF( @days >= 1 AND weekday( S+0 )<=4, 1, 0 )
 +IF( @days >= 2 AND weekday( S+1 )<=4, 1, 0 )
 +IF( @days >= 3 AND weekday( S+2 )<=4, 1, 0 )
 +IF( @days >= 4 AND weekday( S+3 )<=4, 1, 0 )
 +IF( @days >= 5 AND weekday( S+4 )<=4, 1, 0 )
 +IF( @days >= 6 AND weekday( S+5 )<=4, 1, 0 )
 -- days always less than 7 days