匹配范围内可能的日期元素

Question

我很难为日期范围匹配其他情况。最终目标是提取每个组以构建ISO 8601日期格式。

测试用例

May 8th – 14th, 2019
November 25th – December 2nd
November 5th, 2018 – January 13th, 2019
September 17th – 23rd

正则表达式

(w{3,9})s([1-9]|[12]d|3[01])(?:st|nd|rd|th),s(19|20)d{2}s–s(w{3,9})s([1-9]|[12]d|3[01])(?:st|nd|rd|th),s(19|20)d{2}

正则表达式

我希望能够捕获每个组，无论它是否存在。

例如，May 8th – 14th, 2019

Group 1 May
Group 2 8th
Group 3 
Group 4 
Group 5 14th
Group 6 2019

和November 5th, 2018 – January 13th, 2019

Group 1 November
Group 2 5th
Group 3 2018
Group 4 January
Group 5 13th
Group 6 2019

Answer 1

如果组不匹配，要捕获空字符串，一般思路是使用(<characters to match>|)

试试这个：

([A-z]{3,9})s((?:[1-9]|[12]d|3[01])(?:st|nd|rd|th))(?:, (?=19|20))?(d{4}|)s–s([A-z]{3,9}|)s?((?:[1-9]|[12]d|3[01])(?:st|nd|rd|th))(?:, (?=19|20))?(d{4}|)

https://regex101.com/r/4UY0WE/1/

尝试捕获月份(第一组(时，请确保使用[A-z]{3,9}而不是w{3,9}，否则您可能会匹配，例如，23rd而不是月份字符串。

分离出来：

([A-z]{3,9})      # Month ("January")
s
((?:[1-9]|[12]d|3[01])(?:st|nd|rd|th))  # Day of month, including suffix ("23rd")
(?:, (?=19|20))?  # Comma and space, if followed by year
(d{4}|)          # Year
s–s             #
([A-z]{3,9}|)     # same as first line
s?
# same as third to fifth lines:
((?:[1-9]|[12]d|3[01])(?:st|nd|rd|th)) 
(?:, (?=19|20))?
(d{4}|)

Answer 2

这个通过合并一些分组来节省一些空间。

在这里尝试一下

完整正则表达式：

([A-z]{3,9}) ((?:[1-9]|[12]d|3[01])(?:st|nd|rd|th))(?:, ((?:19|20)d{2}))? [–-] ([A-z]{3,9}s)?((?:[1-9]|[12]d|3[01])(?:st|nd|rd|th))(?:, ((?:19|20)d{2}))?

按组分隔(为便于阅读，空格替换为s(：

1. ([A-z]{3,9})
s
2. ((?:[1-9]|[12]d|3[01])(?:st|nd|rd|th))
3. (?:,s((?:19|20)d{2}))?
s[–-]s
4. ([A-z]{3,9}s)?
5. ((?:[1-9]|[12]d|3[01])(?:st|nd|rd|th))
6. (?:,s((?:19|20)d{2}))?

此方法不使用查找，因此对于任何正则表达式引擎通常是安全的。