题目
- [root@OldboyEduoldboy#catip.log
- 10.0.0.1oldboy
- 10.0.0.2oldgirl
- 10.0.0.3bingbing
- 10.0.0.4tingting
- 10.0.0.4oldboy
打印第二列为oldboy行的第一列内容
参考答案
方法一:while循环
@H_502_14@方法二:grep
- catmatch.shwhilereadline
- do
- name=`echo$line|awk'{print$2}'`
- if["$name"="oldboy"];then
- echo$line|awk'{print$1}'
- fi
- done</oldboy/ip.log
@H_502_14@方法三:sed
- [root@showoldboy]#grep-Po"\S+\s(?=oldboy)"ip.txt
- 10.0.0.1
- 10.0.0.4
- [root@showoldboy]#grep-Po"[0-9.]+(?=oldboy)"ip.txt
- 10.0.0.1
- 10.0.0.4
@H_502_14@方法四:awk
- sed-nr'/oldboy/s#(^[^]+).*$#\1#gp'ip.sh
@H_502_14@备注
- awk'$2~/oldboy/{print$1}'ip.sh
今天是每日一题陪伴大家的第92天,期待你的进步。