Parse xml in powershell

I have the following xml:

<?xml version="1.0" encoding="UTF-8"?>
<sections>
<section name="Options">
<item key="HLVersionControlWebServiceURL" value="http://www.personec.no/webservices/HLVersionControl/HLVersionControl.asmx" />
<item key="AltinnWebServiceURL" value="https://www.altinn.no/webservices/DataExchange.asmx" />
<item key="WorkDir" value="F:\Altinn\Work\" />
<item key="CatalogDir" value="F:\Altinn\Work\" />
</section>
<section name="Users">
<item key="1" value="Admin" name="Administrator" fNr="" password="" entsystype="1" entsysid="180967" entsyspassword="" lastLogin="20091111161516" allowra0500="1" allowrf1037="1" allowra01821="1" allowra01822="0" allowrf1015="1" altinnuserpassword="/qwHHYwYinE=" />
</section>
<section name="SchemaTypes">
<item key="RF1037" displayname="Terminoppgave" inputdir="F:\Altinn\Work\" validationschema=".\melding-669-8570.xsd" isSubForm="0" isSignable="0" />
<item key="RA0500" displayname="SSB Lønnsstatistikk" inputdir="C:\Program Files (x86)\Personec\Altinn Monitor\Work\" validationschema=".\melding-868-7612.xsd" isSubForm="0" isSignable="0" />
<item key="RA01821" displayname="SSB Fraværsstatistikk bedrift" inputdir="C:\Program Files (x86)\Personec\Altinn Monitor\Work\" validationschema=".\melding-862-6190.xsd" isSubForm="0" isSignable="0" />
<item key="RF1015" displayname="Årsoppgave m/ LTO" inputdir="C:\Program Files (x86)\Personec\Altinn Monitor\Work\" validationschema=".\melding-210-7928.xsd" orid="210" orversion="7928" isSubForm="0" isSignable="1" />
<item key="RF1015U" displayname="" inputdir="" validationschema=".\melding-1083-7930.xsd" orid="1083" orversion="7930" isSubForm="1" isSignable="1" />
</section>
</sections>

And I need to alter the item key WorkDir in Powershell. When using "regular" xml-read I get to the top sections (options, users, and so on) but not the "item key" nodes within each. How can I edit the value for WorkDir in powershell? (I realize I could just use a dirty string replace but I'd rather do it "properly".

You can load your XML into a LINQ `XDocument` class from PowerShell like this:

[Reflection.Assembly]::LoadWithpartialName("System.Xml.Linq") | Out-Null
$xDoc = [System.Xml.Linq.XDocument]::Parse($myXmlString)

From there you can use the usual LINQ to XML methods to replace the attribute as in [this example][1]. If you prefer you could use the older `XmlDocument` class in a similar way.


[1]:

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You could try this

$xmlFile = "d:\sample.xml"

[xml]$doc = Get-Content $xmlFile
$node = $doc.SelectSingleNode("/sections/section/item[@key='WorkDir']")
$node.Value = "New-Value"
$doc.Save($xmlFile)

You'll still be using some .Net classes and an XPath query to select the node.

This version uses a bit more PowerShell and handles the case of mulitple items with WorkDir keys:

$xml = [xml](Get-Content foo.xml)
$xpath = "/sections/section/item[@key='WorkDir']"
Microsoft.PowerShell.Utility\Select-Xml $xml -XPath $xpath |
Foreach {$_.Node.SetAttribute('value', $pwd)}
$xml.Save("$pwd\bar.xml")

Note, if you have the [PowerShell Community Extensions][1] installed you can use the Format-Xml cmdlet to format the output and save it via Out-File e.g.:

$xml | Format-Xml -AttributesOnNewLine | Out-File bar.xml -enc utf8

OTOH $xml.Save() is easier except that you must remember that it probably doesn't have the correct current dir if you were to specify just the filename. That's why I used "$pwd\bar.xml" in the first example. This is not an issue with PowerShell cmdlets like Out-File.

[1]:

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$invocation = (Get-Variable MyInvocation).Value
$directorypath = Split-Path $invocation.MyCommand.Path


[xml]$userfile = Get-Content $directorypath\MainSetting.xml
$Value = $userfile.GetElementsByTagName("node")

Foreach ($ValueName in $Value)
{
$FinalValue=$ValueName.InnerXML
}