从命令行打印正则表达式的所有匹配项？ [英] Print all matches of a regular expression from the command line?

问题描述

在unix命令行中将所有匹配（每个匹配一行或每行输入一行）打印到正则表达式的最简单方法是什么？请注意，每行输入可能有0个或多个匹配。

我认为必须有一些方法可以通过sed，awk，grep和/或perl，我希望有一个简单的命令行解决方案，以便在未来需要时显示在我的bash历史记录中。

编辑：澄清，我不想打印所有匹配的行，只打印正则表达式的匹配项。例如，一行可能有1000个字符，但与正则表达式只有两个10个字符的匹配。我只对这两个10个字符的匹配感兴趣。

解决方案

假设您只使用非捕获括号，

  perl -wnE'say / yourregex / g'
  

或

  perl -wnE'say for / yourregex / g'
  

示例使用：



  $ echo -ne'fod，food，fad\\\
bar\\\
fooooood\\\
'| perl -wnE'say for / fo * d / g'
 fod 
 food 
 fooooood 
 $ echo -ne'fod，food，fad\\\
bar\\\
fooooood\ n'| perl -wnE'say / fo * d / g'
 fodfood 
 
 fooooood 
  




What's the simplest way to print all matches (either one line per match or one line per line of input) to a regular expression on a unix command line? Note that there may be 0 or more than 1 match per line of input.

I assume there must be some way to do this with sed, awk, grep, and/or perl, and I'm hoping for a simple command line solution so it will show up in my bash history when needed in the future.

EDIT: To clarify, I do not want to print all matching lines, only the matches to the regular expression. For example, a line might have 1000 characters, but there are only two 10-character matches to the regular expression. I'm only interested in those two 10-character matches.

解决方案

Assuming you only use non-capturing parentheses,

perl -wnE'say /yourregex/g'

or

perl -wnE'say for /yourregex/g'

Sample use:

$ echo -ne 'fod,food,fad\nbar\nfooooood\n' | perl -wnE'say for /fo*d/g'
fod
food
fooooood
$ echo -ne 'fod,food,fad\nbar\nfooooood\n' | perl -wnE'say /fo*d/g'
fodfood

fooooood

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