导出到文本文件时比较对象问题

Question

我想要得到如下的输出。

我现在的输出：

Name            Active PrimarySmtpAddress
----            ------ ------------------
DG_Group1       True   [e-mail]
DG_Group2       False  [e-mail]
DG_Group3       True   [e-mail]

我想要的输出：

[e-mail]
[e-mail]
[e-mail]

脚本：

$DistroLists  = Get-DistributionGroup -ResultSize Unlimited
$MessageTrace = Get-MessageTrace -RecipientAddress $DistroLists.PrimarySmtpAddress -startdate (Get-Date).AddDays(-8) -EndDate (Get-Date)
$DistroLists |
Foreach-Object {
$_ | Add-Member -MemberType NoteProperty -Name Active -Value (
$_.PrimarySmtpAddress -in $MessageTrace.RecipientAddress
) -PassThru
} | 
Select-Object Name, Active, PrimarySmtpAddress | Where-Object Active -EQ "FALSE"
Out-File C:output.txt

Answer 1

您可以使用Select-Object -ExpandProperty或ForEach-Object -MemberName从一个或多个管道输入对象中仅获取特定属性的值：

... |Select-Object Name, Active, PrimarySmtpAddress | Where-Object Active -eq $false | ForEach-Object -MemberName PrimarySmtpAddress | Out-File...

Answer 2

您可以跳过前两行，并使用空格字符拆分该行。最后选择下面给出的最后一列：

Get-Content C:Projectslogtext.txt | Select-Object -Skip 2 | ForEach-Object { "$(($_ 
-split 's+',3)[2])" }

[e-mail] 
[e-mail]
[e-mail]