Question

我想删除第一个括号中的数字，并保持括号的其余部分完好无损。

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，但也希望用他们的数字取出第一个括号。

这是正确的吗？

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

Answer 1

您可以使用sed：

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

Answer 2

要在awk中取出带有数字的第一个括号：

$ 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

它删除了第一次出现（所有数字然后）。然后它从Input_file打印行。

下面的日志可能是什么正则表达式？

3 个答案: