以特殊格式表示时间的正则表达式答案

【问题标题】：regular expression to represent time in a special format以特殊格式表示时间的正则表达式
【发布时间】：2010-11-19 13:15:03
【问题描述】：

我有许多包含日期扩展名的文件。例如，今天的文件将被命名为

文件名.20101118

格式为 yyyymmdd。我想为所有过时的文件列出/grep/etc。

【问题讨论】：

标签： regex bash search grep

【解决方案1】：

你的意思是这样的吗？

/filename\.[12][90][0-9][0-9][01][0-9][0-3][0-9]/

请注意，这也匹配无效日期。

或者，更概括地说：

/[a-zA-Z0-9_]+\.[12][90][0-9][0-9][01][0-9][0-3][0-9]/

您可以使用文件名可以包含的任何字符来代替[a-zA-Z0-9_]。

【讨论】：

【解决方案2】：

如果您不想进行任何验证，请尝试此操作。

\.(\d{8})

【讨论】：

【解决方案3】：

注意：我假设您的“文件名”只是字母数字 (ASCII)。

我最好的朋友RegexBuddy 说：

[a-zA-Z0-9]+?\.(19|20)[0-9]{2}(0[1-9]|1[012])(0[1-9]|[12][0-9]|3[01])

匹配项

filename123.20101118
FOOBAR123.19961212

不匹配

FOOBAR.88881201
foobar.20103512
filename.20101235

免责声明：我只是一个快乐的用户，与 RB 没有任何关系

说明

[a-zA-Z0-9]+?\.(19|20)[0-9]{2}(0[1-9]|1[012])(0[1-9]|[12][0-9]|3[01])

Options: case insensitive

Match a single character present in the list below «[a-zA-Z0-9]+?»
   Between one and unlimited times, as few times as possible, expanding as needed (lazy) «+?»
   A character in the range between “a” and “z” «a-z»
   A character in the range between “A” and “Z” «A-Z»
   A character in the range between “0” and “9” «0-9»
Match the character “.” literally «\.»
Match the regular expression below and capture its match into backreference number 1 «(19|20)»
   Match either the regular expression below (attempting the next alternative only if this one fails) «19»
      Match the characters “19” literally «19»
   Or match regular expression number 2 below (the entire group fails if this one fails to match) «20»
      Match the characters “20” literally «20»
Match a single character in the range between “0” and “9” «[0-9]{2}»
   Exactly 2 times «{2}»
Match the regular expression below and capture its match into backreference number 2 «(0[1-9]|1[012])»
   Match either the regular expression below (attempting the next alternative only if this one fails) «0[1-9]»
      Match the character “0” literally «0»
      Match a single character in the range between “1” and “9” «[1-9]»
   Or match regular expression number 2 below (the entire group fails if this one fails to match) «1[012]»
      Match the character “1” literally «1»
      Match a single character present in the list “012” «[012]»
Match the regular expression below and capture its match into backreference number 3 «(0[1-9]|[12][0-9]|3[01])»
   Match either the regular expression below (attempting the next alternative only if this one fails) «0[1-9]»
      Match the character “0” literally «0»
      Match a single character in the range between “1” and “9” «[1-9]»
   Or match regular expression number 2 below (attempting the next alternative only if this one fails) «[12][0-9]»
      Match a single character present in the list “12” «[12]»
      Match a single character in the range between “0” and “9” «[0-9]»
   Or match regular expression number 3 below (the entire group fails if this one fails to match) «3[01]»
      Match the character “3” literally «3»
      Match a single character present in the list “01” «[01]»


Created with RegexBuddy

【讨论】：

【解决方案4】：

最简单的：

find -regex ".*/.*\.[0-9]+$"

更好：

find -regextype posix-basic -regex ".*/.*\.[0-9]\{8\}$"

最好的？：

find -regextype posix-extended -regex ".*/.*\.[0-9]{4}([0][1-9]|1[012])(0[1-9]|[12][0-9]|3[01])$"

最后一个会找到 any 年（我们不想重复 Y2K，现在，是吗？）。并且只有 01-12 范围内的月份和 01-31 范围内的天。当然，它不会验证一个月中的正确天数或闰日。

【讨论】：