bash 删除字符之前的所有内容
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bash remove everything before character
提问by rmartinez
I have the following:
我有以下几点：
netstat -tuna | awk '{ print ; }' | sed -e '1,2d'
which returns:
返回：
10.20.26.143:51697
10.20.26.143:51696
10.20.26.143:51698
10.20.26.143:51693
10.20.26.143:51695
10.20.26.143:51694
:::22
0.0.0.0:68
0.0.0.0:49625
0.0.0.0:5353
:::48727
:::5353
which returns a list of all open ports.. how would i remove everything before the : character? please notice that a few ones have ::: instead of : i just want it to return the ports, remove all text before the last :
它返回所有开放端口的列表.. 我将如何删除 : 字符之前的所有内容？请注意，有几个有 ::: 而不是 : 我只是想让它返回端口，删除最后一个之前的所有文本：
i want it all in one bash command.
我想在一个 bash 命令中完成所有操作。
thanks a lot in advance!
非常感谢！
回答by user unknown
add to your sed-command:
添加到您的 sed 命令：
';s/.*:/:/g'
netstat -tuna | awk '{ print ; }' | sed -e '1,2d;s/.*:/:/g'
should work.
应该管用。
回答by kurumi
Just do it with one awk command. No need to chain too many
只需使用一个 awk 命令即可。无需链接太多
netstat -tuna | awk 'NR>2{ k=split(,a,":");print a[k] }'
or
或者
netstat -tuna | awk 'NR>2{ sub(/.*:/,"",);print }'
or Ruby(1.9+)
或 Ruby(1.9+)
netstat -tuna | ruby -ane 'puts $F[3].sub(/.*:/,"") if $.>2'
if you want unique ports,
如果你想要独特的端口，
netstat -tuna | awk 'NR>2{ sub(/.*:/,"",); uniq[] }END{ for(i in uniq) print i }'
回答by SiegeX
So apparently, you can tuna network, but you can't tuna fish
The following will print all open ports
以下将打印所有打开的端口
netstat -tuna | awk -F':+| +' 'NR>2{print }'
The following will print all uniqueopen ports, i.e. no duplicates sorted in ascending order
以下将打印所有唯一的开放端口，即没有按升序排序的重复项
netstat -tuna | awk -F':+| +' 'NR>2{print }' | sort -nu
