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提示用户数量的脚本；检查他们的主目录中有多少文件并将这些用户的列表发送到标准输出

tag-icon • tag-icon shell-script users
提示用户数量的脚本；检查他们的主目录中有多少文件并将这些用户的列表发送到标准输出

我的家庭作业要求我

  1. 用户在其主目录中存储了太多文件，您希望通知前 5 个违规者。您可能希望为更多或更少的用户再次运行此脚本，因此此选择将提示需要识别的用户数量，检查他们的主目录中有多少文件，并将这些用户的列表发送到标准输出。

这是到目前为止我所掌握的第一部分内容；我只是无法弄清楚#5。

echo " Please enter file name: "
read workingfile 
while true
 do
  echo "1)  Check who is logged into the system and pull up contact information"
  echo "2)  Check who has left a system process running and stop it"
  echo "3)  Check if two users have the same userid, and prompt to change it"
  echo "4)  Get a list of all users on file and list information about them"
  echo "5)   List the top 5 users that have to many files in their home directory and list them"

  echo " "
  echo -n "Choice: "
  read choice
  case "$choice" in

1)
      user=$(who | cut -d' ' -f1 | sort | uniq)
     grep "$user" $workingfile | sed 's/\:/ /g' | sed 's/stu[0-9]*//g'
     break
        ;;

2)
    echo "Please enter a username: "
    read user
    echo $user
    ps -u $user
    echo "Would you like to end the proccess for $user ? (yes or no)"
    read choice2
        if [ $choice2 = "yes" ];
                echo "killall -u USERNAME.”             break 
        else 
            echo "We will not be stopping any background proccesses!"
            break 
        exit
        fi
    ;;

3)
    sed -i '0,/stu5/{s/stu5/stu6/}' myuser.txt 
    sed -i '0,/stu5/{s/stu5/stu4/}' myuser.txt 
        echo "Testing if a user has the same id"
        if [[ $(grep -o "stu[0-9]" $workingfile | uniq -d) != 0 ]]; then
            result=$(grep -o "stu[0-9]" $workingfile | uniq -d)
            echo " We will be replacing the first instance of $result please input what number you'd like to replace it with!: "
            read input
            echo "replacing id..."
            sed -i '0,/stu5/{s/stu5/stu4/}' $workingfile
            cat $workingfile
            break
        else
            echo " There is no result!"
            break
         exit
         fi
    ;;

4) echo "List of all users in file: "
        cat $workingfile | sed 's/\:/ /g' | sed 's/stu[0-9]*//g'
        break

    ;;

答案1

首先，您必须做出一个假设，要么getent passwd实际返回所有有效用户（在某些情况下不会），要么所有用户的主目录都遵循该形式/home/username（这也不是一个很好的假设）。让我们继续第二个假设。然后你可以尝试类似的事情

cd /home
for user in *; do printf "%s\t%s\n" $(find $user -type f | wc -l) $user; done | sort -rn > /tmp/sort.$$
for user in $(head -5 /tmp/sort.$$ | cut -f2); do notify $user; done

您刚刚又做了两个假设。如果用户的文件名称中包含回车符，则计数将关闭。另外，我假设notify这里的命令是某种电子邮件，并且还假设用户的电子邮件与其用户名相同。也许是一个很好的假设，但也并不总是正确的。

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