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，使用空字符串敲击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。没有错误生成，它只显示如下：
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$D
How many passwords do ya want, buddy?:
将Read-Host提示功能移至开始并将密码生成逻辑移至循环中：
function New-RandomPassword {
param()
[int]$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，您将不必递归调用该函数。
我个人将$counter变量转换为参数，然后在内部调用 的脚本中提示用户New-RandomPassword
function New-RandomPassword {
param(
[ValidateRange(1,500)]
[int]$Count
)
$characterPool = "!@#$%^&*0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ_abcdefghijklmnopqrstuvwxyz".ToCharArray()
$i = 0;
while($i++ -lt $count)
{
[pscustomobject]@{
Password = -join ($characterPool | Get-Random -Count 12)
}
}
}
正如mklement0指出的那样，您还可以为Read-Host调用分配一个表达式作为参数的默认值：
param(
[ValidateRange(1,500)]
[int]$Count = $(Read-Host 'How many passwords do ya want, buddy?')
)
我个人认为这是一种反模式，宁愿坚持默认1或标记参数Mandatory：
# If the caller omits the argument, 1 password is returned
param(
[ValidateRange(1,500)]
[int]$Count = 1
)
要么
# 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
)
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