计算两个日期之间有多少 n.month 问：start_date + n.months >= end_dates，变量n是什么？

如何解决计算两个日期之间有多少 n.month 问：start_date + n.months >= end_dates，变量n是什么？

我想按月计算两个日期之间的距离：

问：start_date + n.months >= end_dates，变量n是什么？

start_date = Date.parse('2021-01-31')
end_date = Date.parse('2021-02-28')

## start_date + 1.months = 2021-02-28
## The answer 1 month,which is more clearable for human using
## Don't want to find the answer like 28 days or 0.93 month. (30 day a month)

首先我试图让每个月是 30.days，但答案会在某些特殊日期上有一些偏差。每个月的日期都不同，End of Feb month 上的日期始终是问题所在。

然后我尝试安装像 date_diff、time_difference... 这样的 gem，但是没有任何方法可以做到这一点，大部分输出是 28 天而不是 1 个月。

对于简单的方法，我可以轻松地进行迭代循环来找到 n，例如：

start_date = Date.parse('2021-01-31')
end_date = Date.parse('2021-02-28')

def calc_diff(start_date,end_date)
  n = 0
  while start_date + n.months < end_date
     n += 1
  end
  n
end

有没有更好的方法可以找到两个日期之间的 n 个月，而不是使用循环？

谢谢。

解决方法

我对问题的理解与以下示例一致。我已经计算了 Date 对象 date1 和 date2 之间的差异，其中 date2 >= date1。

require 'date'

def months_between(date1,date2)
  12*(date2.yr - date1.yr) + date2.mon - date1.mon + date2.day > date1.day ? 1 : 0
end

months_between Date.new(2020,1,22),Date.new(2020,3,21) #=> 2
months_between Date.new(2020,22) #=> 2 
months_between Date.new(2020,23) #=> 3

months_between Date.new(2020,Date.new(2021,21) #=> 14
months_between Date.new(2020,22) #=> 14
months_between Date.new(2020,23) #=> 15

,

# find minimum n so that `start_date + n.months >= end_dates`
def calc_diff(start_date,end_date)
 diff = (end_date.yday - start_date.yday) / 30
 return diff if start_date + diff.months >= end_date
 diff + 1
end

calc_diff(Date.parse('2021-01-31'),Date.parse('2021-02-28')) # 1
calc_diff(Date.parse('2021-01-31'),Date.parse('2021-04-30')) # 3
calc_diff(Date.parse('2021-01-31'),Date.parse('2021-05-31')) # 4
calc_diff(Date.parse('2021-02-01'),Date.parse('2021-06-01')) # 4
calc_diff(Date.parse('2021-02-01'),Date.parse('2021-06-02')) # 5

,

感谢@Cary 和@Lam 的回答。

这是我找到 n month 的答案。

# try to find the minimum n month between start_date and target_date
def calc_diff(start_date,target_date)
  months_diff = (target_date.year * 12 + target_date.month) - (start_date.year * 12 + start_date.month)

  ## need to check the end of month because some special case
  ## start date: 2020-01-31 ; end date 2020-06-30
  ## the minimum n month must be 5
  ## another special case of Feb must consider (test case 15)

  if start_date.day > target_date.day && !((start_date == start_date.end_of_month || target_date.month == 2) && (target_date == target_date.end_of_month))
    months_diff = months_diff - 1
  end
  puts months_diff # it will show the minimum whole n month

  # the target_date will between inside
  # (start_date + months_diff.months) <= target_date < (start_date + (months_diff + 1).months) 
  (start_date + months_diff.months)..(start_date + (months_diff + 1).months) 
end

测试用例：

## test case 1
## 6/15 - 7/15  => n = 5
calc_diff(Date.parse('2020-01-15'),Date.parse('2020-06-19'))

## test case 2
## 7/15 - 8/15  => n = 6
calc_diff(Date.parse('2020-01-15'),Date.parse('2020-07-15')) 

## test case 3
## 5/15 - 6/15  => n = 4
calc_diff(Date.parse('2020-01-15'),Date.parse('2020-06-01')) 

## test case 4 (special case)
## 6/30 - 7/31  => n = 5
calc_diff(Date.parse('2020-01-31'),Date.parse('2020-06-30'))

## test case 5
## 7/30 - 8/30  => n = 4
calc_diff(Date.parse('2020-04-30'),Date.parse('2020-07-31'))

## test case 6
## 6/30 - 7/30  => n = 2
calc_diff(Date.parse('2020-04-30'),Date.parse('2020-06-30'))  

## test case 7
## 5/31 - 6/30  => n = 4
calc_diff(Date.parse('2020-01-31'),Date.parse('2020-05-31'))

## test case 8
## 2/29 - 3/31  => n = 1
calc_diff(Date.parse('2020-01-31'),Date.parse('2020-02-29'))

## test case 9
## 6/29 - 7/29  => n = 4
calc_diff(Date.parse('2020-02-29'),Date.parse('2020-06-30'))

## test case 10
## 7/29 - 8/29  => n = 5
calc_diff(Date.parse('2020-02-29'),Date.parse('2020-07-31'))

## test case 11
## 1/31 - 2/29  => n = 0
calc_diff(Date.parse('2020-01-31'),Date.parse('2020-02-28'))

## test case 12
## 2/29 - 3/31  => n = 1
calc_diff(Date.parse('2020-01-31'),Date.parse('2020-03-01'))

## test case 13
## 1/17 - 2/17  => n = 0
calc_diff(Date.parse('2020-01-17'),Date.parse('2020-01-17'))

## test case 14
## 1/17 - 2/17  => n = 0
calc_diff(Date.parse('2020-01-17'),Date.parse('2020-01-18'))

## test case 15 (special case)
## 1/30 - 2/29  => n = 1
calc_diff(Date.parse('2019-12-30'),Date.parse('2020-02-28'))

## test case 16
## 2/29 - 3/30  => n = 2
calc_diff(Date.parse('2019-12-30'),Date.parse('2020-02-29'))