bash中的Perl oneliner：从复杂的regexp打印匹配项

Question

I have this complex regex 我有这个复杂的正则表达式

/"_outV":([0-9]+),"_inV":([0-9]+),"_label":"([a-z\/]+)",/

and I need to parse a file (which is all on one single line) and output only the matched groups like 我需要解析一个文件（全部在一行上），然后仅输出匹配的组，例如

print $1 $2 $3

Currently the only almost working onliner is 目前只差一点工作 onliner是

perl -pe 'while(m/"_outV":([0-9]+)\,"_inV":([0-9]+)\,"_label":"([a-z\/]+)\"\,/g){print "$1 $2 $3\n";}'

But it ends up echoing also the entire file at the end, after the matches. 但是最终在匹配之后，它还会在最后回显整个文件。 How do I fix this? 我该如何解决？

I though that removing the -p option would make the trick, but it doesn't. 我虽然删除了-p选项可以解决问题，但事实并非如此。

Answer 1

Looks good to me. 对我来说看上去很好。 You need to replace the -p with -n and here is why . 您需要将-p 替换为-n ，这就是原因。

A few finer points: 一些要点：

No need to backslash those , and " . 无需反斜杠那些,和" 。
You can conveniently replace [0-9] with \\d . 您可以方便地将[0-9]替换为\\d 。
By using a different delimiter for the regex you won't need to escape the / . 通过为正则表达式使用其他定界符，您将不需要转义/ 。

End result optimized 最终结果优化

perl -ne 'print "$1 $2 $3\n" while m{"_outV":(\d+),"_inV":(\d+),"_label":"([a-z/]+)",}g'