Bash密码强度校验脚本如何同时支持-f读文件与直接传密码参数

实现原理 #

Bash的getopts处理完所有带-前缀的选项后，内置变量$OPTIND会指向第一个非选项参数的位置，我们只需要将已经处理完的选项参数从位置参数列表中移除，就能直接读取用户传入的无选项密码参数。

具体修改步骤 #

你只需要在原代码的while getopts循环结束后，新增如下代码即可：

# 移除所有已经处理完成的选项参数
shift $((OPTIND - 1))

# 判断是否已经通过-f选项获取到密码，未获取则读取位置参数作为密码
if [ -z "$password" ]; then
    # 校验用户是否传入了密码参数，未传入则提示用法
    if [ $# -eq 0 ]; then
        echo "用法错误：请要么通过 -f 选项指定存储密码的文件，要么直接传入密码作为参数"
        echo "用法示例："
        echo "$0 -f passwd.txt"
        echo "$0 MyTestPass123"
        exit 2
    fi
    password="$1"
fi

额外优化提示 #

1. 原代码中初始化requirements数组的语句是requirements=(foo bar)，数组初始长度只有2，但后续代码用到了索引2的位置，会导致赋值逻辑异常，建议将该初始化语句修改为requirements=()即可正常运行。
2. 原代码中读取密码文件的语句password=cat $OPTARG``建议修改为password=$(cat "$OPTARG")，避免文件路径包含空格时读取失败。

完整修改后代码 #

#!/bin/bash
while getopts ":f:" option; do
      case $option in
             f) password=$(cat "$OPTARG") ;;
      esac
done

# 新增功能代码开始
shift $((OPTIND - 1))
if [ -z "$password" ]; then
    if [ $# -eq 0 ]; then
        echo "用法错误：请要么通过 -f 选项指定存储密码的文件，要么直接传入密码作为参数"
        echo "用法示例："
        echo "$0 -f passwd.txt"
        echo "$0 MyTestPass123"
        exit 2
    fi
    password="$1"
fi
# 新增功能代码结束

#evaluating how much chars the password has
password_length=${#password}
#counter for checking in how much sections the password meets the rquirements
count=0
#修正数组初始化语句
requirements=()
#Checking if password includes minimum of 10 characters
if [ $password_length -ge 10 ];
then
    requirements[0]="Correct"
else
    requirements[0]="Incorrect password syntax. The password length must includes minimum of 10 characters"
fi
#checkig if the password includes both alphabet and number
if [[ "$password" == *[a-zA-Z]* && "$password" == *[0-9]* ]]
then
    requirements[1]="Correct"
    
else
    requirements[1]="Incorrect password syntax. The password must includes both alphabet and number"
    
fi
#checking if password includes both the small and capital case letters.
if [[ "$password" == *[A-Z]* && "$password" == *[a-z]* ]];
then
    requirements[2]="Correct"
else
    requirements[2]="Incorrect password syntax. The password must includes both the small and capital case letters"
    
fi
#checking whether the password is according to the requirements or not 
#if yes count will equal to 3 at the end
for i in "${requirements[@]}"
do
    if [[ $i == "Correct" ]];
    then
        let count++
    fi
done
#if the counter count is equal to 3 print the password in light green color and return exit 0
if [[ $count -eq 3 ]];
then
    echo -e "\e[92m$password"
# sleep - user has the time to see that the password's syntax is correct 
    sleep 3
    exit 0 
#if the count is not equal to 3 print the password in reg color and return exit 1
else
    echo -e "\e[91m$password"
    for i in "${requirements[@]}"
    do
        if [[ $i != "Correct" ]];
        then
            echo $i        
        fi
    done
# sleep - user has the time to see that the password's syntax is incorrect and the reasons for that
    sleep 6
    exit 1
fi

内容的提问来源于stack exchange，提问作者Avichai Abutbul