我的 New-RandomPassword 功能突然不起作用

Question

为了好玩，我决定创建一个函数，该函数会根据用户决定他们想要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 块有一个 $characterPool 变量，其中字母数字和特殊字符作为潜在字符。 Process 块询问用户想要多少个密码并创建许多随机的 12 个字符的密码。当我第一次创建该函数时，它运行良好。现在，无论我请求多少密码，都只会创建两个密码，然后它会重新提出问题。之后，使用空字符串按回车键或添加另一个数字会导致打印出一个密码。

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
    }

我希望这会生成我请求的密码数量。没有产生错误，它只是显示如下：

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提示函数移到开头，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)
        }
    }
}

这样你就不必递归调用函数了。

---

我会亲自将$counter 变量转换为参数，然后在内部调用 New-RandomPassword 的脚本中提示用户

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)
        }
    }
}

---

作为mklement0 pointed out，您还可以使用Read-Host 调用作为参数的默认值来分配表达式：

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，并且宁愿坚持默认为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
)