Question

我不想根据第6列和第7列打印重复的行。sort -u似乎无济于事

cat / tmp / testing：-

-rwxrwxr-x. 1 root root  52662693 Feb 27 13:11 /home/something/bin/proxy_exec
-rwxrwxr-x. 1 root root  27441394 Feb 27 13:12 /home/something/bin/keychain_exec
-rwxrwxr-x. 1 root root  45570820 Feb 27 13:11 /home/something/bin/wallnut_exec
-rwxrwxr-x. 1 root root  10942993 Feb 27 13:12 /home/something/bin/log_exec
-rwxrwxr-x. 1 root root 137922408 Apr 16 03:43 /home/something/bin/android_exec

当我尝试cat /tmp/testing | sort -u -k 6,6 -k 7,7时，我得到：-

-rwxrwxr-x. 1 root root 137922408 Apr 16 03:43 /home/something/bin/android_exec
-rwxrwxr-x. 1 root root  52662693 Feb 27 13:11 /home/something/bin/proxy_exec

期望的输出如下，因为这是唯一根据月和日期列与其他文件不同的文件

-rwxrwxr-x. 1 root root 137922408 Apr 16 03:43 /home/something/bin/android_exec

Answer 1

在gnu sed上尝试

sed -E '/^\s*(\S+\s+){5}Feb\s+27/d' testing

在gnu awk上尝试

awk 'NR==1{a=$6$7;next} a!=$6$7{print}' testing

Answer 2

[not]使用awk打印基于第6列和第7列的重复行，您可以：

$ awk '
++seen[$6,$7]==1 {          # count seen instances
    keep[$6,$7]=$0          # keep first seen ones       
}
END {                       # in the end
    for(i in seen)
        if(seen[i]==1)      # the ones seen only once
            print keep[i]   # get printed
}' file                     # from file or pipe your ls to the awk

给定输入的输出：

-rwxrwxr-x. 1 root root 137922408 Apr 16 03:43 /home/something/bin/android_exec

通知：关于解析ls输出的所有标准警告仍然适用。

我不想根据第6列和第7列打印重复的行

2 个答案: