ruby - 如何提取 ruby 中的一部分线？

Question

我有话要说

line = "start running at Sat April 1 07:30:37 2017"

我想提取

"Sat April 1 07:30:37 2017"

我试过了...

line = "start running at Sat April 1 07:30:37 2017"
if (line =~ /start running at/)
   line.split("start running at ").last
end

...但是还有其他方法吗？

Answer 1

这是一种从任意字符串中提取代表给定格式时间的子字符串的方法。我假设字符串中至多有一个这样的子字符串。

require 'time'

R = /
    (?:#{Date::ABBR_DAYNAMES.join('|')})\s
              # match day name abbreviation in non-capture group. space
    (?:#{Date::MONTHNAMES[1,12].join('|')})\s
              # match month name in non-capture group, space
    \d{1,2}\s # match one or two digits, space
    \d{2}:    # match two digits, colon
    \d{2}:    # match two digits, colon
    \d{2}\s   # match two digits, space
    \d{4}     # match 4 digits
    (?!\d)    # do not match digit (negative lookahead)
    /x        # free-spacing regex def mode
  # /
  #  (?:Sun|Mon|Tue|Wed|Thu|Fri|Sat)\s
  #   (?:January|February|March|...|November|December)\s
  # \d{1,2}\s
  # \d{2}:
  # \d{2}:
  # \d{2}\s
  # \d{4}
  # (?!\d)
  # /x 

def extract_time(str)
  s = str[R]
  return nil if s.nil?
  (DateTime.strptime(s, "%a %B %e %H:%M:%S %Y") rescue nil) ? s : nil
end

str = "start eating breakfast at Sat April 1 07:30:37 2017"
extract_time(str)
  #=> "Sat April 1 07:30:37 2017" 

str = "go back to sleep at Cat April 1 07:30:37 2017"
extract_time(str)
  #=> nil

或者，如果与 R 匹配，但是 Time#strptime引发异常(意味着 s 不是给定时间格式的有效时间)可以引发异常以建议用户。