Question

我正在使用Powershell v3在Windows Server 2012上测试我的应用程序的部署脚本。该脚本在Win Server 2008 R2和Win 7上使用Powershell v2运行正常。我现在遇到的问题是我无法访问通过-ArgumentList传递的XML变量的属性。

我已经能够在一个简单的脚本中使用Powershell v3在Win 7和Win Server 2012上重现该问题，该脚本没有我的主脚本所做的任何SharePoint，IIS，misc。

脚本（我想我从一个我现在找不到的类似问题中借用了这个）：

$xml = [xml] (Get-Content (Get-Item (".\input.xml")))
$foobar = $xml.Stuff.FooBars.Foobar 

$ScriptBlock = {        
    $foobar = $args[0]

    write-host "Inside the job..."
    write-host ("Foobar     : "+$foobar)
    write-host ("Foobar.Foo : "+$foobar.Foo)
}

write-host "Outside the job..."
write-host ("Foobar: "+$foobar)
write-host ("Foobar.Foo : "+$foobar.Foo)

Start-Job -ScriptBlock $ScriptBlock -ArgumentList $foobar | Out-Null        
While (Get-Job -State "Running") { Start-Sleep 2 }               

write-host ("Jobs Completed.")    
Get-Job | Receive-Job          
Remove-Job *

输入XML：

<?xml version="1.0"?>
<Stuff xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
  <FooBars>
    <FooBar>
      <Foo>ThisIsAFoo</Foo>
     <Bar>ThisIsABar</Bar>
    </FooBar>
  </FooBars> 
</Stuff>

Powershell v2的输出：

PS C:\Powershell3Issues> .\demo.ps1
Outside the job...
Foobar: System.Xml.XmlElement
Foobar.Foo : ThisIsAFoo
Jobs Completed.
Inside the job...
Foobar     : System.Collections.ArrayList System.Collections.ArrayList
Foobar.Foo : ThisIsAFoo

Powershell v3的输出：

PS C:\Powershell3Issues> .\demo.ps1
Outside the job...
Foobar: System.Xml.XmlElement
Foobar.Foo : ThisIsAFoo
Jobs Completed.
Inside the job...
Foobar     : System.Collections.ArrayList System.Collections.ArrayList
Foobar.Foo :

注意缺少的Foobar.Foo值。

我也尝试过v3中的$ using语法，但它也做了同样的事情。

Answer 1

尝试此操作以指定在以下位置运行作业的PowerShell版本：

Start-Job -ScriptBlock $ScriptBlock -ArgumentList $foobar -PSVersion 2.0

Answer 2

我正在使用PS 3.0，它确实改变了数据类型。我修改了你的脚本来看看：

$xml = [xml] (Get-Content .\input.xml)
$foobar = $xml.Stuff.FooBars.Foobar 
"Foobar Outside type = $($foobar.Gettype())"

$ScriptBlock = {        
    $foobar = $args[0]
    "Foobar Inside type = $($foobar.Gettype())"
}

Start-Job -ScriptBlock $ScriptBlock -ArgumentList $foobar | Out-Null
While (Get-Job -State "Running") { Start-Sleep 2 }     
Get-Job | Receive-Job

我得到的输出是：

Foobar Outside type = System.Xml.XmlElement
Foobar Inside type = System.Collections.ArrayList System.Collections.ArrayList

我会一直看着，看看我发现了什么。

<强>更新

当我使用命令powershell -version 2.0在PS 2.0中运行脚本时，我收到错误消息：

Method invocation failed because [Deserialized.System.Xml.XmlElement] doesn't contain a method named 'Gettype'.
+ CategoryInfo          : InvalidOperation: (Gettype:String) [], RuntimeException
+ FullyQualifiedErrorId : MethodNotFound
+ PSComputerName        : localhost

它从System.Xml.XmlElement更改为Deserialized.System.Xml.XmlElement，对吗？我会继续看，看看我发现了什么。

解决方法：

解决方法可能是传递一个对象而不是传递一个字符串。

$xml_path = 'C:\input.xml' $sb = { $xml = [xml](Get-Content $args[0]) ... } Start-Job -ScriptBlock $sb -Args $xml_path

我已经完成了搜索。当我深入研究事物时，我的头疼。

Answer 3

问题是您正在尝试序列化XMLElement类型的对象，该对象不会序列化。解决方法是克隆XMLElement并将其包装在新的XMLDocument中。

cls

$XMLDocument = [xml]@"
<?xml version="1.0"?>
<Stuff xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
  <FooBars>
    <FooBar>
      <Foo>ThisIsAFoo1</Foo>
      <Bar>ThisIsABar2</Bar>
    </FooBar>
    <FooBar>
      <Foo>ThisIsAFoo2</Foo>
      <Bar>ThisIsABar2</Bar>
    </FooBar>
  </FooBars> 
</Stuff>
"@

Select-Xml -Xml $XMLDocument -XPath 'Stuff/FooBars/FooBar' | foreach {  
  $SelectedNode = $_.Node
  #$SelectedNode is now of type System.Xml.XmlElement which does not serialise
  #but System.Xml.XmlDocument will serialise so wrap a clone of $SelectedNode in a new XMLDocument 
  #then pass that as the argument
  $SerializationWrapper = New-Object System.Xml.XmlDocument  
  $SerializationWrapper.AppendChild($SerializationWrapper.ImportNode($SelectedNode, $true)) | Out-Null
  Start-Job -ScriptBlock {
    param($xml)
    Write-Output "Start Job"    
    'The type of deserialise object $xml is: ' + $xml.getType()    
    $sw = New-Object system.io.stringwriter 
    $writer = New-Object system.xml.xmltextwriter($sw) 
    $writer.Formatting = [System.xml.formatting]::Indented 
    $xml.WriteContentTo($writer) 
    $sw.ToString()  
    Write-Output "Finish Job"    
    Write-Output ""    
  } -ArgumentList $SerializationWrapper
}

Get-Job | Wait-Job | Receive-Job

从下面的输出可以看出，生成了2个作业，并传递了一个包装好的FooBar元素进行格式化。

Id     Name            PSJobTypeName   State         HasMoreData     Location             Command                  
--     ----            -------------   -----         -----------     --------             -------                  
520    Job520          BackgroundJob   Running       True            localhost            ...                      
522    Job522          BackgroundJob   Running       True            localhost            ...                      
Start Job
The type of deserialise object $xml is: xml
<FooBar>
  <Foo>ThisIsAFoo1</Foo>
  <Bar>ThisIsABar2</Bar>
</FooBar>
Finish Job

Start Job
The type of deserialise object $xml is: xml
<FooBar>
  <Foo>ThisIsAFoo2</Foo>
  <Bar>ThisIsABar2</Bar>
</FooBar>
Finish Job

Start-Job将XML对象传递给-ArgumentList，它在Powershell V2＆amp; V3

3 个答案: