Linux lsof 需要帮助解析输出

Question

我在 linux 系统上使用以下命令：

lsof -i -n | egrep '\<ssh\>'|awk '{print $8,$9}'

它会产生如下输出：

192.168.199.52:ssh->192.168.199.254:17598 (ESTABLISHED)
192.168.199.52:ssh->192.168.199.254:17598 (ESTABLISHED)
192.168.199.52:56448->69.168.130.22:ssh (ESTABLISHED)
192.168.199.52:56449->69.168.130.22:ssh (ESTABLISHED)
192.168.199.52:56454->69.168.130.22:ssh (ESTABLISHED)
192.168.199.52:56458->69.168.130.22:ssh (ESTABLISHED)
192.168.199.52:56460->69.168.130.22:ssh (ESTABLISHED)
192.168.199.52:56468->69.168.130.22:ssh (ESTABLISHED)
192.168.199.52:ssh->192.168.199.254:56671 (ESTABLISHED)
192.168.199.52:ssh->192.168.199.254:56671 (ESTABLISHED)
192.168.199.52:ssh->192.168.199.254:56672 (ESTABLISHED)

我只想提取左侧的 IP 地址和“->”字段右侧的 IP 地址。如何轻松提取这两个文件并将它们重新组合成以下格式：

192.168.199.52->192.168.199.254

Answer 1

类似：

lsof -i -n | awk '$9 ~ /:ssh(-|$)/{ gsub(/:[^-]*/, "", $9); print $9 }'

或者用 8 美元代替 9 美元。

awk 命令详情：

$9 ~ /:ssh(-|$)/ {           # when ":ssh" is at the end of field 9 or
                             # followed by an hyphen
    gsub(/:[^-]*/, "", $9);  # remove all the semi-colon followed by characters that
                             # are not an hyphen from the field 9
    print $9                 # and print it
}