02-Shell编程-Linux用户和组

Shell脚本实现编程管理Linux用户和组脚本，编写思路如下：

---

（1）脚本支持创建普通用户。

（2）支持创建多个用户或者列表用户添加。

（3）支持Linux系统用户删除。

（4）支持Linux系统组删除。

（5）支持对某个用户修改密码。

相关代码如下

#!/bin/bash
#2024年4月30日
#auto manager linux user

########################
# 以下是一个用于自动管理 Linux 用户的 Shell 脚本。

USR="$*"

# 首先检查是否以 root 用户身份运行此脚本
if [ $UID -ne 0 ];then
    echo -e "\033[32m-----------------\033[0m"
    echo -e "\033[32mThe script must be executed using the root user.\033[0m"
    exit 1
fi

# 定义添加用户的函数
add_user(){
    read -p "Please enter the user name you need to create? " USR
    for USR in $USR
    do
        # 检查用户是否存在，如果不存在，则创建用户并设置密码
        id $USR
        if [ $? -ne 0 ];then
            useradd -s /bin/bash $USR -d /home/$USR
            echo ${USR}_123456|passwd --stdin $USR
            if [ $? -eq 0 ];then
                echo -e "\033[32m-----------------\033[0m"
                echo -e "\033[32mThe $USR user created successfully\033[0m"
                echo -e "User,Password"
                echo -e "$USR,${USR}_123"
                echo
                tail -n 5 /etc/passwd
            fi
        else
            echo -e "\033[32m-----------------\033[0m"
            echo -e "\033[32mThis $USR user already exists, please exit\033[0m"
            exit 1
        fi
    done
}

# 定义从用户列表添加用户的函数
add_user_list(){
    # 从用户列表文件中读取用户名，并逐个添加用户
    if [ $# -eq 1 ];then
        while read USR
        do
            useradd -s /bin/bash $USR -d /home/$USR
            echo ${USR}_123456|passwd --stdin $USR
            if [ $? -eq 0 ];then
                echo -e "\033[32m-----------------\033[0m"
                echo -e "\033[32mThe $USR user created successfully\033[0m"
                echo -e "User,Password"
                echo -e "$USR,${USR}_123"
                echo
                tail -n 5 /etc/passwd
            fi
        done < $1
    else
        echo -e "\033[32m-----------------\033[0m"
        echo -e "\033[32mThe user list file must be entered. The reference content format is as follows:\033[0m"
        echo "cyly1"
        echo "cyly2"
        echo "cyly3"
        echo "cyly4"
        echo "......"
    fi
}

# 定义删除用户的函数
remove_user(){
    for USR in $USR
    do
        # 逐个删除用户及其相关文件
        userdel -r $USR
        if [ $? -eq 0 ];then
            echo -e "\033[32m-----------------\033[0m"
            echo -e "\033[32mThe $USR user delete successfully\033[0m"
            echo
            tail -n 5 /etc/passwd
        fi
    done
}

# 定义删除用户组的函数
remove_group(){
    for USR in $USR
    do
        # 逐个删除用户组
        groupdel $USR
        if [ $? -eq 0 ];then
            echo -e "\033[32m-----------------\033[0m"
            echo -e "\033[32mThe $USR group delete successfully\033[0m"
            echo
            tail -n 5 /etc/passwd
        else	
            grep "$USR" /etc/group
            if [ $? -eq 0 ];then
                echo -e "\033[32m-----------------\033[0m"
                echo -e "\033[32mThe $USR group delete failed, cannot remove the primary group of user $USR\033[0m"
                read -p "Are you sure you want to delete the $USR user? yes or no " INPUT
                if [ $INPUT == "y" -o $INPUT == "Y" -o $INPUT == "yes" -o $INPUT == "YES" ];then
                    userdel -r $USR
                    groupdel $USR >>/dev/null 2>&1
                    echo -e "\033[32m-----------------\033[0m"
                    echo -e "\033[32mThe $USR user delete successfully\033[0m"
                    echo -e "\033[32mThe $USR group delete successfully\033[0m"
                fi
            fi
        fi
    done
}

# 定义修改用户密码的函数
change_user_passwd(){
    read -p "Please enter your user name and new password: username password: " INPUT
    if [ $(echo $INPUT|sed 's/ /\n/g'|wc -l) -eq 2 ];then
        USR=$(echo $INPUT|awk '{print $1}')
        PAS=$(echo $INPUT|awk '{print $2}')
        for USR in $USR
        do
            echo $PAS|passwd --stdin $USR		
            if [ $? -eq 0 ];then
                echo -e "\033[32m-----------------\033[0m"
                echo -e "\033[32mThe password of $USR user was modified successfully\033[0m"
                echo -e "User,Password"
                echo -e "$USR,$PAS"
                echo
            fi
        done
    fi
}

# 根据用户输入执行相应的操作
case $1 in 
    1)
        add_user
        ;;
    2)
        add_user_list $2
        ;;
    3)
        remove_user
        ;;
    4)
        remove_group
        ;;
    5)
        change_user_passwd
        ;;
    *)
        echo "---------------------------------------------"
        echo -e "\033[34mWelcome to system user management scripts:\033[0m"
        echo -e "\033[32m1) add user\033[0m"
        echo -e "\033[32m2) add_user_list\033[0m"
        echo -e "\033[32m3) remove_user\033[0m"
        echo -e "\033[32m4) remove_group\033[0m"
        echo -e "\033[32m5) change_user_passwd\033[0m"
        echo -e "\033[32mUsage:{/bin/sh $0 1|2|3|4|5|help}\033[0m"
        echo "---------------------------------------------"
esac

使用教程（会的不用看哦）

我用的MobaXterm，先创建一个shell名称

将上面的代码复制进去

给他赋值权限

chmod o+x user.sh

RUN

./user.sh

这里我们输入 1不起作用，我们要用下列命令来执行1-5 或者help

具体意思看顶部需求

./user.sh 1