问题描述
在几个具有不同命名架构的不同域中工作。因此，我正在编写一个进入每个域的脚本，并检查它们的组成员身份。
Working in a few different domains, that have different naming schema's. So I'm writing a script that goes into each domain, and checks on their group membership.
脚本要做的第一件事是询问用户的姓氏。然后我使用 Get-ADUser 选择 samaccountname 并将其绑定到变量（ samaccountname 是 Get-ADPrincipalGroupMembership -Identity 参数接受的唯一名称）。
First thing the script does is ask for the user's last name. Then I use Get-ADUser to select the samaccountname and tie it to a variable (samaccountname is the only name that Get-ADPrincipalGroupMembership -Identity parameter accepts).
但是当我使用 -Identity 变量运行脚本时，找不到用户。如果我手动输入-它确实找到了用户。
But when I run the script using the variable for -Identity it doesn't find the user. If I type it in manually - it does find the user.
这里是代码：
$surname = Read-Host "Users Last Name"
$fullname = Get-ADUser -filter * | Where-Object {$_.surname -eq $surname} |
select samaccountname | Format-Table -HideTableHeaders | Out-String
Get-ADPrincipalGroupMembership -Identity $fullname | select name |
Format-Table -HideTableHeaders
我得到的错误表明变量是一个字符串，并且它是要搜索的正确用户，但错误提示它找不到该用户。
The error I get shows that the variable is a string, and it is the correct user that it's searching for, but the error says it can't find that user.
推荐答案
Format-* cmdlet用于向用户显示数据。当需要/打算对数据进行进一步处理时，请勿使用它们。
Format-* cmdlets were made for displaying data to a user. Do not use them when further processing of the data is required/intended.
更改
... | select samaccountname | Format-Table -HideTableHeaders | Out-String
至
... | select -Expand samaccountname -First 1
，您的问题就会消失。
这篇关于Get-ADPrincipalGroupMembership-身份不接受变量的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！