powershell Exchange 2003：查询 ActiveDirectory 的 activesync、office 和描述？

Question

我使用 powershell 来提取 Exchange 2003 用户的列表，并根据用户的邮箱“LegacyDN”，我希望使用 LDAP 查询来提取他们的 activesyncenabled 属性 (msExchOmaAdminWirelessEnable)、描述和 Office； 然而，它似乎并不适合我。 有任何想法吗？

这是我查询 AD 属性的方式：

$allusers = get-wmiobject -class exchange_mailbox -namespace root\microsoftexchangev2 -computername srv02 | select mailboxdisplayname, legacyDN 

foreach ($user in $allusers) 
{
$obj = new-object psobject 
$exchangever = "2003" 

$tmp = [adsi]("LDAP://" + $user.legacydn) 

$obj | add-member -membertype noteproperty -name "Display Name" -value $user.mailboxdisplayname
$obj | add-member -membertype noteproperty -name "Office" -value $tmp.physicaldeliveryofficename 
$obj | add-member -membertype noteproperty -name "Description" -value $tmp.description 
$obj | add-member -membertype noteproperty -name "Activesync" -value $tmp.msExchOmaAdminWirelessEnable 

$allusersemail += $obj 

}

$alluseremail | export-csv \\srv02\logs\test.csv 

Answer 1

这是一种方法：

$search = System.DirectoryServices.DirectorySearcher

$search.filter = "(&(homeMDB=*)(objectcategory=user)(objectclass=user))"

$props = "mailNickName","displayName"," physicaldeliveryofficename","描述","msExchOmaAdminWirelessEnable"

$props | foreach { [void]$search.PropertiesToLoad.Add($_) }

$users = $search.findAll()

foreach($u in $users)

{

$DisplayName = @{name="DisplayName";表达式={$u .properties.item("DisplayName")}}

$Office = @{name="Office";表达式={$u.properties.item("physicaldeliveryofficename")}}

$Description = @{name="Description";表达式={$u.properties.item("描述")}}

$Activesync = @{name="Activesync";表达式={$u.properties.item("msExchOmaAdminWirelessEnable")}}

$u | 选择 $DisplayName、$Office、$Description、$Activesync

}