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  • awk进阶

    1.[4@izuf60jasqavbxb9efockpz ~]$ echo $PATH | awk 'BEGIN{FS=":"}''{print $0}'

    /usr/local/bin:/usr/bin:/usr/local/sbin:/usr/sbin:/home/4/.local/bin:/home/4/bin:/home/16116374/test

    [4@izuf60jasqavbxb9efockpz ~]$ echo $PATH | awk 'BEGIN{RS=":"}''{print $0}'

    /usr/local/bin
    /usr/bin
    /usr/local/sbin
    /usr/sbin
    /home/4/.local/bin
    /home/4/bin
    /home/4/test

    RS会根据制定的分隔符换行

    2.[4@izuf60jasqavbxb9efockpz ~]$ echo $PATH | awk 'BEGIN{FS=":"}''{print $0}'

    /usr/local/bin:/usr/bin:/usr/local/sbin:/usr/sbin:/home/4/.local/bin:/home/4/bin:/home/4/test

    3.[4@izuf60jasqavbxb9efockpz ~]$echo $PATH | awk 'BEGIN{FS=":"}''{print $1}'

    /usr/local/bin

    科学计算与词典使用

    4.awk 'BEGIN{print 3*7}'

    21

    5.echo | awk '{print 3*7}'

    21

    6.awk 'BEGIN{printf "%.3f ",3/7}'

    0.429

    7.awk 'BEGIN{printf "%.f ",3/7}'

    0

    8.awk 'BEGIN{printf "%.1f ",3/7}'

    0.4

    9.seq 10 2 20

    10
    12
    14
    16
    18
    20

    步数为2

    10.seq 10 2 20 | awk '{sum+=$1}{print sum}'

    10
    22
    36
    52
    70
    90

    11.seq 10 2 20 | awk '{sum+=$1}{print sum,NR}'

    10 1
    22 2
    36 3
    52 4
    70 5
    90 6

    12.seq 10 2 20 | awk '{sum+=$1}{print sum/NR}'

    10
    11
    12
    13
    14
    15

    13.seq 10 20 | awk '$1>19'

    20

    大于19

    14. seq 10 2 20 | awk '$1>14 {data[NR]=$1}{for (k in data) print data[k]}'

    16
    16
    18
    16
    18
    20

    15.seq 10 2 20 | awk '$1>14 {data[NR]=$1}END{for (k in data) print data[k]}'

    16
    18
    20

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  • 原文地址:https://www.cnblogs.com/godlove/p/10697900.html
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