7

I'm looking to use PowerShell to alter XML. I haven't been able to copy XML using XPath. I can load the XML, but I can't figure out how to list the XML with an XPath and create another XML file with what I retrieve. 

$doc = new-object "System.Xml.XmlDocument"
$doc.Load("XmlPath.xml")

Also, how would you add the removed XML to another XML file?

Anthony Mastrean
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Bruce227
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    Just a note - you can simply load xml document like this: $doc = [xml](get-content xmlpath.xml) or $doc = [xml](gc xmlpath.xml) – stej Nov 12 '09 at 06:18

2 Answers2

15

If you're using PowerShell 2.0 you can use the new Select-Xml cmdlet to select xml based on an XPath expression e.g.:

$xml = '<doc><books><book title="foo"/></books></doc>'
$xml | Select-Xml '//book'
Node    Path          Pattern
----    ----          -------
book    InputStream   //book

To remove nodes:

PS> $xml =[xml]'<doc><books><book title="foo"/><book title="bar"/></books></doc>'
PS> $xml | Select-Xml -XPath '//book' | 
        Foreach {$_.Node.ParentNode.RemoveChild($_.Node)}

title
-----
foo
bar

PS> $xml.OuterXml
<doc><books></books></doc>

Then to save to file:

$xml.Save("$pwd\foo.xml")
Get-Content foo.xml
<doc>
  <books>
  </books>
</doc>
Keith Hill
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  • when i just do the $xml | select-xml -xpath 'xpath' i don't get any output. what is being piped into the foreach. i'm not following what the select is doing. is there an easy way to replace values at a certain xpath location? thanks – Bruce227 Nov 12 '09 at 15:09
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    Do you happen to be using the PowerShell Community Extensions? If it is PSCX then I believe it is an XPathNavigator but if you're using the PowerShell 2.0 Select-Xml it is a custom object they create that has a Node property. – Keith Hill Nov 12 '09 at 15:12
  • no, i'm not using the powershell community extensions. so are you saying that $xml | select-xml -xpath '//Configuration/ConfigXML/Configuration/DiscoveryHandlers' should work? well it doesn't for me. the xpath i'm putting doesn't seem to be working. any ideas? thanks for all the help. – Bruce227 Nov 12 '09 at 16:26
  • The sample code that Keith has posted works for me, are you sure your XML and XPath are correct maybe use a tool such as http://pgfearo.googlepages.com/home to check your XPath against your XML. –  Nov 12 '09 at 19:42
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    Bruce, does your XML doc use XML namespaces? If so then the XPath query has to be tweaked and you have to provide namespace mappings to Select-Xml via the -Namespace parameter. If that is the case, I could provide you an example that uses an XML namespace. – Keith Hill Nov 12 '09 at 20:27
  • +1 Bruce. And I found this useful: http://huddledmasses.org/xpath-and-namespaces-in-powershell/ – Shashi Penumarthy Jun 05 '12 at 15:41
2

Load Linq Xml assemblies:

[System.Reflection.Assembly]::LoadWithPartialName("System.Xml.Linq")
[System.Reflection.Assembly]::LoadWithPartialName("System.Xml.XPath")

Load your xml (Note, you can use ::Load("file") instead of ::Parse(...) to load from file:

$xml = [System.Xml.Linq.XDocument]::Parse("<root> <row>Hey</row> <row>you</row> </root>")

Modify (in this case Remove the first row:

[System.Xml.XPath.Extensions]::XPathSelectElement($xml, "//row").Remove()

Save to file:

$xml.Save("MyXml.xml")

Using System.Xml (instead of System.Xml.Linq):

$doc = new-object "System.Xml.XmlDocument"
$doc.Load("MyXml_int.xml")

$node = $doc.SelectSingleNode("//row");
$node.ParentNode.RemoveChild($node)

$doc.Save("MyXml_out.xml")
Nestor
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  • well i'm doing this part of server loading applications and altering the config files for those applications. i don't want to have to load another assembly if possible. is there not a way to do this with the normal powershell? thanks – Bruce227 Nov 12 '09 at 03:51
  • Bruce227, I've added the example using System.Xml. – Nestor Nov 12 '09 at 05:21
  • hi nestor, i tried your system.xml and it didn't work. is the "//row" the xpath value? if so i'm just getting a null when i am using the xpath i have generated. thanks – Bruce227 Nov 12 '09 at 15:16
  • "//row" works with the Xml "Heyyou". You need to change that to whatever xpath fits your xml. I tried my example in my machine and it worked. – Nestor Nov 12 '09 at 15:20
  • hi nester, the xpath examples i have looked at don't seem to have the // just the / for root and then the path. – Bruce227 Nov 12 '09 at 16:28