Question

#!/bin/bash

password=$1

 if [[ $# -gt 1 || $1 = "-h" ]]; then
            echo 'Usage: pw | pw -h | pw password'
            echo "Note:   Valid passwords must be between 8-16 characters long.
            contain at least 1 digit
            contain at least 1 lowercase letter
            contain at least 1 uppcase letter
            contain one of @ # $ % & * + - ="
 fi
if [[ $# -lt 1 ]]; then
    regex='0-9A-Z@*+#$%&a-z'
    password=$( cat /dev/urandom | tr -dc '0-9A-Z@*+#$%&a-z' | head -c $(( 8 + $RANDOM % 8 )))
    check=$( echo $password | grep -o ['@*+#$%&'] | wc -m )
    while [[ $check  -gt 2 || $check -lt 1 ]]
            do
            password=$( cat /dev/urandom | tr -dc '0-9A-Z@*+#$%&a-z' | head -c $(( 8 + $RANDOM % 8 )))
            check=$( echo $password | grep -o ['@*+#$%&'] | wc -m )
    done
    echo $password
 fi
if [[ $# = 1 ]]; then
    password=$1
    echo "$password" | grep [A-Z][a-z]
    if [[ $? = 1 ]]; then

    echo "Password must contain at least one upper or lowercase characer"
            exit 2
    fi

    echo "$password" | grep [0-9]
    if [[ $? = 1 ]]; then

    echo "Password must contain at least one digit"
            exit 3
    fi

    if [[ `echo "$password" | grep -o ['@#$%&*+-'] 2>/dev/null` -gt 2 ]]; then
            echo "You must only use one special character"
            exit 4
    fi

    if [[ `echo "$password" | grep -o ['\@\#\$\%\&\*\+\-'] 2>/dev/null` -lt 1 ]]; then

    echo "Password must contain at least one special character"
            exit 5
    fi
    woc=$( echo "$password" | wc -m )
    if [[ $woc -lt 8 || $woc -gt 16 ]]; then

    echo "Password must be between 8 and 16 characters"
            exit 6
    fi
 echo "$1 is a valid password"
fi

此脚本的第一部分工作正常，并生成一个8-16个字符长的随机密码，只有一个特殊字符。问题出现在第40行和第45行，两者都给出了同样的错误＆＃34;语法错误：期待操作数＆＃34; [特殊字符]＆＃34;。我尝试使用$？创建变量并调用它们？变量来检查输出等等。不幸的是，即使脚本运行正确，此错误也会保持一致。

Answer 1

我修复了代码，这是一个简单的错误。我忘了包含wc -m。

if [[ `echo "$password" | grep [\@\#\$\%\&\*\+\-] | wc -m` -lt 2 ]]

使用grep对包含特殊字符的变量进行BASH脚本语法错误

1 个答案: