#!/bin/bash users=$(cut -d: -f1 /etc/passwd) for user in $users do if [[ $(getent passwd $user | cut -d: -f3) -ge 1000 ]] then if [[ $(chage -l $user | grep "Password expires" | grep "密码过期时间" | cut -d: -f2 | tr -d ' ') == "never" || $(chage -l $user | grep "Password expires" | grep "密码过期时间" | cut -d: -f2 | tr -d ' ') == "从不" ]] then echo "$user: Password never expires" else expire_date=$(date -d "$(chage -l $user | grep "Password expires" | cut -d: -f2 | tr -d ' ')" +%s) current_date=$(date +%s) if [[ $expire_date -lt $current_date ]] then echo "$(hostname -I | awk '{print $1}'), $user: Password has expired" else echo "$user: Password will expire on $(chage -l $user | grep "Password expires" | cut -d: -f2 | tr -d ' ')" fi fi else account_status=$(passwd -S $user | awk '{print $2}') if [[ "$account_status" == "L" ]]; then echo "$(hostname -I | awk '{print $1}'), User $user: account locked!" fi fi done这段代码去掉筛选非系统用户

可以将第二个if语句中的条件改为：

if [[ $(getent passwd $user | cut -d: -f3) -ge 1000 && $(getent passwd $user | cut -d: -f7) != "/sbin/nologin" && $(getent passwd $user | cut -d: -f7) != "/usr/sbin/nologin" ]]

这样就可以排除掉系统用户了。

相关问题

#!/bin/bash users=$(cut -d: -f1 /etc/passwd) for user in $users do if [[ $(getent passwd $user | cut -d: -f3) -ge 1000 ]] then if [[ $(chage -l $user | grep "Password expires" | cut -d: -f2 | tr -d ' ') == "never" ]] then echo "$user: Password never expires" else expire_date=$(date -d "$(chage -l $user | grep "Password expires" | cut -d: -f2 | tr -d ' ')" +%s) current_date=$(date +%s) if [[ $expire_date -lt $current_date ]] then echo "$(hostname -I | awk '{print $1}'),$user: Password has expired" else echo "$user: Password will expire on $(chage -l $user | grep "Password expires" | cut -d: -f2 | tr -d ' ')" fi fi else echo "" account_status=$(passwd -S $user | awk '{print $2}') if [[ "$account_status" == "L" ]]; then echo "$(hostname -I | awk '{print $1}'),User $user: account locked!" fi done语法问题

在第27行缺少一个 `fi` 关键字来结束第20行的 `if` 语句。同时，第24行的 `echo` 命令中，应该把逗号放在双引号内，否则会被视为两个参数。正确的脚本如下：

#!/bin/bash

users=$(cut -d: -f1 /etc/passwd)

for user in $users
do
    if [[ $(getent passwd $user | cut -d: -f3) -ge 1000 ]]
    then
        if [[ $(chage -l $user | grep "Password expires" | cut -d: -f2 | tr -d ' ') == "never" ]]
        then
            echo "$user: Password never expires"
        else
            expire_date=$(date -d "$(chage -l $user | grep "Password expires" | cut -d: -f2 | tr -d ' ')" +%s)
            current_date=$(date +%s)
            if [[ $expire_date -lt $current_date ]]
            then
                echo "$(hostname -I | awk '{print $1}'), $user: Password has expired"
            else
                echo "$user: Password will expire on $(chage -l $user | grep "Password expires" | cut -d: -f2 | tr -d ' ')"
            fi
        fi
    else
        account_status=$(passwd -S $user | awk '{print $2}')
        if [[ "$account_status" == "L" ]]; then
            echo "$(hostname -I | awk '{print $1}'), User $user: account locked!"
        fi
    fi
done

用shell写一个脚本，编写程序，将/etc/ passwd的第一栏取出，而且每一栏都以一行字符串“The1 account is "root””显示，其中1表示行数。

可以使用如下的shell脚本实现：

#!/bin/bash

# 读取/etc/passwd文件的第一栏
users=$(cut -d: -f1 /etc/passwd)

# 遍历每个用户名，并输出对应的信息
count=1
for user in $users
do
    echo "The$count account is $user"
    count=$((count+1))
done

这个脚本首先使用`cut`命令获取/etc/passwd文件的第一栏（即用户名），然后使用循环遍历每个用户名，并输出对应的信息。其中，`$count`表示行数，`$user`表示当前用户名。注意要使用`$((count+1))`进行变量的自增操作。