如果未提供值，如何从cmdlet省略powershell参数? [英] How to omit powershell parameter from cmdlet if value not provided?

问题描述

我正在尝试执行以下Powershell命令，以使用名称，容器等参数在交换服务器上创建新的地址列表.

I'm trying to execute a below Powershell command to create a new address list on exchange server with parameters like Name, Container, etc.

Container是可选的输入/参数，如果未提供cmdlet的值，如何从cmdlet中忽略它?

Container is an optional input/parameter, how do I omit it from cmdlet if its value is not provided?

我尝试使用IF条件句，但似乎不起作用.这里有帮助吗?

I tried with IF conditionals but but does not seems working. Any help here?

New-AddressList -Name -Container \test MyAddressList5 -ConditionalStateOrProvince maha -IncludedRecipients MailboxUsers

推荐答案

您可以通过 hashtable 传递所需的参数及其相应的值.添加If/Else条件以包括属性.像这样:

You can pass needed parameters with their corresponding values via hashtable. Add If/Else conditions to include properties. Like so:

$Container  = '\test MyAddressList5'

$Parameters = @{}
$Parameters.Add('ConditionalStateOrProvince','maha')
$Parameters.Add('IncludedRecipients','MailboxUsers')
if($Container){$Parameters.Add('Container',$Container)}

New-AddressList @Parameters

此外，当您需要包含Switch参数时，只需传递$True.像这样:

Also, when you need to include Switch parameter just pass $True. Like so:

$Parameters.Add('SomeSwitchParameter',$True)

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