给日志花费时间排序 + awk去掉最后几个字母 + awk练习题
#去掉/npmsbak/oss-emulator/store/apaas-storage/9cf/img/20200824后面的日期
awk '{sub(/.{8}$/,"")}1' /opt/0629500/awkimgdir.txt > /opt/0629500/goodmvcmd.sh
(15条消息) awk去掉每行最后n个字符_weixin_30289831的博客-CSDN博客
统计1天24小时每个小时某个类型接口的访问量
cat access_2023-02-02.log|grep 'biz' |awk '{split($2,a,":");b[a[2]]++} END{for(i in b) print(i,b[i])}'|sort
cat access_2023-02-02.log|grep 'biz' |awk '{split($2,a,":");b[a[2]]++} END{for(i in b) printf("%10s %5d\n",i,b[i])}'|sort
打印日志花费时间大于21ms的日志
1 12ms
2 25ms
3 45ms
4 21ms
cat time.txt |awk '{if(int( substr($2,0,length($2)-2) )>21) print $0}'
用一个例子来演示会更加清晰