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Linux下的排序与计算

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Linux下的排序与计算 
reboot   system boot  3.10.0-327.el7.x Wed Oct 26 15:12 - 22:43  (07:30)    
root     :0           :0               Wed Oct  5 05:01 - 05:31  (00:29)    
(unknown :0           :0               Wed Oct  5 05:01 - 05:01  (00:00)    
reboot   system boot  3.10.0-327.el7.x Tue Oct  4 23:01 - 05:31  (06:29)    
root     :0           :0               Wed Oct  5 04:56 - 04:58  (00:01)    
(unknown :0           :0               Wed Oct  5 04:56 - 04:56  (00:00)    
reboot   system boot  3.10.0-327.el7.x Tue Oct  4 22:55 - 04:58  (06:02)

任何人都可以告诉我如何在 Linux 中编写一个命令来计算每个用户的登录次数并计算所花费的总时间。将其投影到显示此内容的屏幕上。

User root logged in a total of ?? times with a total of mins/hours logged
User unknown logged in a total of ?? times with a total of mins/hours logged.

答案1

reboot   system boot  3.10.0-327.el7.x Wed Oct 26 15:12 - 22:43  (07:30)    
root     :0           :0               Wed Oct  5 05:01 - 05:31  (00:29)    
(unknown :0           :0               Wed Oct  5 05:01 - 05:01  (00:00)    
reboot   system boot  3.10.0-327.el7.x Tue Oct  4 23:01 - 05:31  (06:29)    
root     :0           :0               Wed Oct  5 04:56 - 04:58  (00:01)    
(unknown :0           :0               Wed Oct  5 04:56 - 04:56  (00:00)    
reboot   system boot  3.10.0-327.el7.x Tue Oct  4 22:55 - 04:58  (06:02)

考虑到上述详细信息位于文件中：

for i in `awk '{print $1}' test_login |sort|uniq`;
 do
         count=`grep -c $i test_login`
        sum=0;
        for j in `grep $i test_login|awk '{print $NF}'|cut -c 2-6`;
         do
                 var=`echo $j | awk -F: '{print ($1 * 60) + $2}'`;
                 sum=$((sum+var));
        done
        ((hour=$sum/60))
        ((min=$sum-$hour*60))
        echo "User $i logged in a total of $count times with a total of $hour:$min logged"
  done

test_login ：包含所有用户日志的文件。

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