以下logs的正则表达式可能是什么？

Question

我想删除第一个括号中的数字，并保持所有其余的括号完好无损。

Mar 17 00:03:13   %ASA-5-106100: access-list   permitted tcp  10.252.0.165(50811) ->  172.19.26.33(4902) 
Mar 17 00:03:16   %ASA-5-106100: access-list   permitted tcp  10.252.0.166(54563) ->  172.19.26.33(4902) 
Mar 17 00:03:28   %ASA-5-106100: access-list   permitted tcp  10.252.0.222(38071) ->  172.19.26.33(4902) 
Mar 17 00:03:41   %ASA-5-106100: access-list   permitted tcp  10.252.0.222(38074) ->  172.19.26.33(4902) 
Mar 17 00:03:45   %ASA-5-106100: access-list   permitted tcp  10.252.0.221(17868) ->  172.19.26.33(4902) 
Mar 17 00:03:58   %ASA-5-106100: access-list   permitted tcp  10.252.0.166(54572) ->  172.19.26.33(4902) 
Mar 17 00:03:58   %ASA-5-106100: access-list   permitted tcp  10.252.0.166(54573) ->  172.19.26.33(4902) 
Mar 17 00:03:58   %ASA-5-106100: access-list   permitted tcp  10.252.0.166(54574) ->  172.19.26.33(4902) 
Mar 17 00:04:14   %ASA-5-106100: access-list   permitted tcp  10.252.0.165(50826) ->  172.19.26.33(4902) 
Mar 17 00:04:16   %ASA-5-106100: access-list   permitted tcp  10.252.0.166(54580) ->  172.19.26.33(4902) 
Mar 17 00:04:28   %ASA-5-106100: access-list   permitted tcp  10.252.0.222(38088) ->  172.19.26.33(4902) 
Mar 17 00:04:45   %ASA-5-106100: access-list   permitted tcp  10.252.0.221(17881) ->  172.19.26.33(4902) 

我想保持(4902)完好无损，但也想去掉第一个括号和它们的数字。

这样使用正确吗？

awk '{sub('()'..... dst'()'," dst")}1'

Answer 1

您可以使用sed

sed 's/([0-9]*)//' logfile

Answer 2

在awk中去掉第一个括号和它们的数字：

$ awk '{sub(/\([^)]*\)/,"")}1' foo
Mar 17 00:03:13   %ASA-5-106100: access-list   permitted tcp  10.252.0.165 ->  172.19.26.33(4902) 
Mar 17 00:03:16   %ASA-5-106100: access-list   permitted tcp  10.252.0.166 ->  172.19.26.33(4902) 

Answer 3

@Bhavik:试试：

awk '{sub(/\([0-9]+\)/,"");print}'   Input_file

它删除(所有数字，然后)的第一个匹配项。然后，它打印Input_file中的代码行。