[英]How to format the Out-GridView display using calculated property?
下面的 Powershell 腳本非常適合顯示 Azure 用戶屬性,除了需要格式化以刪除公司域名的許可證列。
這是腳本。
#Import Module
If (!(Get-Module "*MSOnline*")) {Import-Module MSOnline}
If (!(Get-Module "*Exchange*")) {Import-Module $((Get-ChildItem -Path $($env:LOCALAPPDATA + "\Apps\2.0\") -Filter Microsoft.Exchange.Management.ExoPowershellModule.dll -Recurse).FullName | ?{ $_ -notmatch "_none_" } | select -First 1)}
#Set admin UPN
$UPN = 'Global.Admin@domain.com'
#This connects to Azure Active Directory & Exchange Online
Connect-MsolService
$EXOSession = New-ExoPSSession -UserPrincipalName $UPN
Import-PSSession $EXOSession -DisableNameChecking -AllowClobber
$startsWith = @(
'Test'
'Sync_'
)
$endsWith = @(
'365'
'\$'
'svc'
'Sync'
'user'
)
$pattern = '^({0})|({1})$' -f $($startsWith -join '|'), $($endsWith -join '|')
# Member Outputs for Microsoft.Online.Administration.User based on https://docs.microsoft.com/en-us/powershell/module/msonline/get-msoluser?view=azureadps-1.0
$allUsers = @()
$allUsers = Get-MsolUser -All -EnabledFilter EnabledOnly | Where-Object {
($_.UserPrincipalName -notmatch $pattern) -and
($_.UserPrincipalName -notlike '*#EXT#*') -and
($_.DisplayName -notmatch 'Admin|Calendar|Room|Prod|Account|Fax|Team|Office|Test|User')
} | Select-Object FirstName,
LastName,
UserPrincipalName,
@{Label = 'SMTP Address(es)';
Expression = {
If (( $_.UserPrincipalName -match 'onmicrosoft.com$')) {
($_.proxyAddresses | Where-Object { ($_ -like 'SMTP*') -and ($_ -like '*onmicrosoft.com') }) -replace 'smtp:' -join ';'
} Else {
($_.proxyAddresses | Where-Object { ($_ -like 'SMTP*') -and ($_ -notlike '*onmicrosoft.com') }) -replace 'smtp:' -join ';'
}
}
},
AlternateEmailAddresses,
UsageLocation,
isLicensed,
Licenses,
@{Label = 'License(s)';
Expression = {
($_.Licenses | ForEach-Object { ($_.AccountSkuId | Where-Object { $_.AccountSkuId -like 'CorpFinanceLtd*' }) -replace 'CorpFinanceLtd:' } ) -join ';'
}
},
PasswordNeverExpires,
BlockCredential
$allUsers | Out-GridView
我相信這部分需要一些修改,但不知道該怎么做?
@{Label = 'License(s)';
Expression = {
($_.Licenses | ForEach-Object { ($_.AccountSkuId | Where-Object { $_.AccountSkuId -like 'CorpFinanceLtd*' }) -replace 'CorpFinanceLtd:' } ) -join ';'
}
},
默認許可證列顯示許可證信息,如: {CorpFinanceLtd:MCOPSTNC, CorpFinanceLtd:ENTERPRISEPREMIUM, CorpFinanceLtd:EMSPREMIUM, CorpFinanceLtd:RIGHTSMANAGEMENT_ADHOC…}
那么如何刪除 CorpFinanceLtd: 前綴部分,從而產生更有意義的結果?
您將$_.AccountSkuId
的值作為管道輸入傳遞,但繼續在結果對象上尋找AccountSkuId
屬性 - 這不起作用,因為它只是一個字符串。
您可以將表達式簡化如下:
@{
Label = 'License(s)';
Expression = {
($_.Licenses | ForEach-Object { $_.AccountSkuId -replace '^CorpFinanceLtd:' } ) -join ';'
}
},
CorpFinanceLtd:
前面的^
:確保正則表達式引擎只匹配字符串的開頭,不再需要通配符過濾器
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.