组合查询(WQL子选择查询- Powershell - SCCM)
组合查询(WQL子选择查询- Powershell - SCCM)
提问于 2019-06-06 02:50:39
现在我正在使用两个不同的查询并比较结果对象，但是我更喜欢一个唯一的查询来完成所有需要的查询，因为我希望直接在SCCM中使用它，而不仅仅是PowerShell。
(第一个查询创建一个包含所有安装了某个x64软件的计算机的对象，第二个对象创建一个包含所有未安装某个x86软件的计算机的查询)
比较这两个对象，我可以得到我需要的列表(哪台机器在两个对象中)
但是，我如何将这两个查询组合在一起，使其成为一个查询呢？
如下所示：
所有有SMS_G_System_ADD_REMOVE_PROGRAMS_64.DisplayName = "SOFTWARE1“但没有SMS_G_System_ADD_REMOVE_PROGRAMS.DisplayName = "SOFTWARE2”的计算机
$SiteCode = "x"
$SiteServer = "x"
$query = @"
select * from SMS_R_System
inner join SMS_G_System_ADD_REMOVE_PROGRAMS_64
on SMS_G_System_ADD_REMOVE_PROGRAMS_64.ResourceId = SMS_R_System.ResourceId
where SMS_G_System_ADD_REMOVE_PROGRAMS_64.DisplayName = "SOFTWARE1"
"@
$postes_xx = (Get-WmiObject -namespace root\sms\site_$SiteCode -computer $SiteServer -query $query).SMS_R_SYSTEM.Name
$query = @"
select SMS_R_SYSTEM.ResourceID,SMS_R_SYSTEM.ResourceType,SMS_R_SYSTEM.Name,SMS_R_SYSTEM.SMSUniqueIdentifier,SMS_R_SYSTEM.ResourceDomainORWorkgroup,SMS_R_SYSTEM.Client
from SMS_R_System
inner join SMS_G_System_COMPUTER_SYSTEM
on SMS_G_System_COMPUTER_SYSTEM.ResourceID = SMS_R_System.ResourceId
where SMS_G_System_COMPUTER_SYSTEM.Name
not in (select distinct SMS_G_System_COMPUTER_SYSTEM.Name
from SMS_R_System
inner join SMS_G_System_COMPUTER_SYSTEM
on SMS_G_System_COMPUTER_SYSTEM.ResourceID = SMS_R_System.ResourceId
inner join SMS_G_System_ADD_REMOVE_PROGRAMS
on SMS_G_System_ADD_REMOVE_PROGRAMS.ResourceID = SMS_R_System.ResourceId
where SMS_G_System_ADD_REMOVE_PROGRAMS.DisplayName = "SOFTWARE2" )
"@
$postes_32x = Get-WmiObject -namespace root\sms\site_$SiteCode -computer $SiteServer -query $query | select -ExpandProperty name
Compare-Object $postes_xx $postes_x32 -IncludeEqual -ExcludeDifferent回答 2
Stack Overflow用户
回答已采纳
发布于 2019-06-06 10:15:14
似乎不需要使用SMS_G_System_Computer_System类。下面是WQL的一个简单版本，它应该可以满足您的需求。
select SMS_R_System.Name
from SMS_R_System
where SMS_R_System.ResourceID in
(select SMS_G_System_ADD_REMOVE_PROGRAMS_64.ResourceID
from SMS_G_System_ADD_REMOVE_PROGRAMS_64
where SMS_G_System_ADD_REMOVE_PROGRAMS_64.DisplayName = "SOFTWARE1")
and SMS_R_System.ResourceID not in
(select SMS_G_System_ADD_REMOVE_PROGRAMS.ResourceID
from SMS_G_System_ADD_REMOVE_PROGRAMS
where SMS_G_System_ADD_REMOVE_PROGRAMS.DisplayName = "SOFTWARE2")希望我的回答能对你有所帮助，并期待你的反馈。
致以最好的问候，雷
Stack Overflow用户
发布于 2019-06-06 06:31:35
看起来您可以只包含所有的连接，然后将where语句与and结合起来。您必须调整select语句以包含您关心的列。
$query = @"
select * from SMS_R_System
inner join SMS_G_System_ADD_REMOVE_PROGRAMS_64
on SMS_G_System_ADD_REMOVE_PROGRAMS_64.ResourceId = SMS_R_System.ResourceId
inner join SMS_G_System_COMPUTER_SYSTEM
on SMS_G_System_COMPUTER_SYSTEM.ResourceID = SMS_R_System.ResourceId
where (SMS_G_System_ADD_REMOVE_PROGRAMS_64.DisplayName = "SOFTWARE1")
and (SMS_G_System_COMPUTER_SYSTEM.Name
not in (select distinct SMS_G_System_COMPUTER_SYSTEM.Name
from SMS_R_System
inner join SMS_G_System_COMPUTER_SYSTEM
on SMS_G_System_COMPUTER_SYSTEM.ResourceID = SMS_R_System.ResourceId
inner join SMS_G_System_ADD_REMOVE_PROGRAMS
on SMS_G_System_ADD_REMOVE_PROGRAMS.ResourceID = SMS_R_System.ResourceId
where SMS_G_System_ADD_REMOVE_PROGRAMS.DisplayName = "SOFTWARE2" ))
"@页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/56466389
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