Question

为了娱乐，我决定创建一个函数，该函数将根据用户决定的数量生成一些随机密码。我正在使用Begin和Process块来运行脚本。 Begin块具有$ characterPool变量，其中字母数字和特殊字符为潜在字符。 Process块询问用户需要多少个密码，并创建那么多随机的12个字符的密码。当我第一次创建函数时，它工作正常。现在，无论我请求多少个密码，都只会创建两个，然后重新提出问题。然后，使用空字符串按Enter或添加另一个数字将导致输出一个密码。

Begin
    {
        $characterPool = "!@#$%^&*0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ_abcdefghijklmnopqrstuvwxyz".ToCharArray()
    }
    Process
    {
        [pscustomobject]@{
            Password = -join ($characterPool | Get-Random -Count 12)
            }
        $counter = Read-Host -Prompt "How many passwords do ya want, buddy?"
        1..$counter | New-RandomPassword
    }

我希望这会生成我要求的密码数量。不会产生任何错误，它只会显示如下：

Begin
    {
        $characterPool = "!@#$%^&*0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ_abcdefghijklmnopqrstuvwxyz".ToCharArray()
    }
    Process
    {
        [pscustomobject]@{
            Password = -join ($characterPool | Get-Random -Count 12)
            }
        $counter = Read-Host -Prompt "How many passwords do ya want, buddy?"
        1..$counter | New-RandomPassword
    }

How many passwords do ya want, buddy?: 5
Password    
--------    
1z@p8Jgl52yU
1#K8z2@iA!&o
How many passwords do ya want, buddy?: 
Pw!C#d2xTjMu
How many passwords do ya want, buddy?: 
fo*Oca9HbRQr
How many passwords do ya want, buddy?: 
omzCRpwqdOfM
How many passwords do ya want, buddy?: 
u5lMN!kjhzfe
How many passwords do ya want, buddy?: 6
vLaxHq945K$D
How many passwords do ya want, buddy?:

Answer 1

将Read-Host提示功能移至开始并将密码生成逻辑移入循环：

function New-RandomPassword {
    param()

    [int]$counter = Read-Host -Prompt "How many passwords do ya want, buddy?"
    $characterPool = "!@#$%^&*0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ_abcdefghijklmnopqrstuvwxyz".ToCharArray()
    $i = 0;
    while($i++ -lt $counter)
    {
        [pscustomobject]@{
            Password = -join ($characterPool | Get-Random -Count 12)
        }
    }
}

这样，您将不必递归调用函数。

我个人将$counter变量转换为参数，然后在脚本中提示用户，该脚本在内部调用 New-RandomPassword

function New-RandomPassword {
    param(
        [ValidateRange(1,500)]
        [int]$Count
    )

    $characterPool = "!@#$%^&*0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ_abcdefghijklmnopqrstuvwxyz".ToCharArray()
    $i = 0;
    while($i++ -lt $count)
    {
        [pscustomobject]@{
            Password = -join ($characterPool | Get-Random -Count 12)
        }
    }
}

与mklement0 pointed out一样，您也可以使用Read-Host调用将表达式指定为参数的默认值：

param(
    [ValidateRange(1,500)]
    [int]$Count = $(Read-Host 'How many passwords do ya want, buddy?')
)

我个人认为这是一种反模式，宁愿坚持默认为1或标记参数Mandatory：

# If the caller omits the argument, 1 password is returned
param(
    [ValidateRange(1,500)]
    [int]$Count = 1
)

或

# Caller _must_ supply a parameter argument (the default console host will prompt the user if not, but depends on the host application)
param(
    [Parameter(Mandatory = $true)]
    [ValidateRange(1,500)]
    [int]$Count
)

我的New-RandomPassword函数突然不起作用

1 个答案: