Selecting attributes in xml using xpath in powershell

Question

I am trying to use powershell and XPath to select the name attribute shown in the below xml example.

     $xml_peoples= $file.SelectNodes("//people") 
     foreach ($person in $xml_peoples){
            echo $person.attributes
            #echo $person.attributes.name
     }

Above is the code im running to try and get the name, but it doesn't seem to work. Any suggestions?

<peoples>
    <person name='James'>
        <device>
            <id>james1</id>
            <ip>192.192.192.192</ip>
        </device>
    </person>
</peoples>

Thanks in advance!

Is it a typo? Your xpath has "people" by you have no XML nodes with that name... Is it supposed to be "person" instead? — Andy Arismendi
– Andy Arismendi, Commented Jul 11, 2013 at 1:19
Please specify "doesn't seem to work." What actually happened? I think I know what you expected to happen, but it wouldn't hurt to specify that too. — LarsH
– LarsH, Commented Jul 11, 2013 at 1:19

Ansgar Wiechers · Accepted Answer · 2013-07-11 09:03:07Z

37

These two lines should suffice:

[xml]$xml = Get-Content 'C:\path\to\your.xml'
$xml.selectNodes('//person') | select Name

answered Jul 11, 2013 at 9:03

Ansgar Wiechers

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3 Comments

Sum1sAdmin Over a year ago

how can we cast all xml files in a directory as xml objects in Powershell? as in $path = get-childitem -filter *.xml (foreach $file in $path) {...#what goes here}

Ansgar Wiechers Over a year ago

If you have a new question: please post a new question. Don't followup on someone else's question in comments.

Sum1sAdmin Over a year ago

sorry, I've posted as a new question stackoverflow.com/questions/42561059/…

CommonToast · Accepted Answer · 2014-08-12 07:40:54Z

26

How about one line?

Select-XML -path "pathtoxml" -xpath "//person/@name"

answered Aug 12, 2014 at 7:40

CommonToast

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2 Comments

John Neuhaus Over a year ago

That would be perfect, if it wasn't for Select-XML's atrocious handling of namespaces. You'd need to add a namespace parameter (with a dummy namespace name if it uses the default xmlns like most) for every element in the path. And it's no longer a one-liner (or at least a readable one) if you don't know the namespace value ahead of time.

Cheeso Feb 16 at 5:25

To get the version out of a .csproj file: (Select-XML -path MyProject.csproj -xpath "/Project/PropertyGroup/Version/text()").node.Value

Adi Inbar · Accepted Answer · 2013-07-11 01:58:04Z

16

I'm not sure what $hub is, and you started your code from the middle so it's not clear if you properly set $file to an XmlDocument object, but I think this is what you want:

[System.Xml.XmlDocument]$file = new-object System.Xml.XmlDocument
$file.load(<path to XML file>)
$xml_peoples= $file.SelectNodes("/peoples/person")
foreach ($person in $xml_peoples) {
  echo $person.name
}

answered Jul 11, 2013 at 1:58

Adi Inbar

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Comments

John Neuhaus · Accepted Answer · 2017-08-22 16:43:26Z

12

For anyone that has to work around Select-Xml's garbage namespace handling, here's a one-liner that doesn't care, as long as you know the direct path:

([xml](Get-Content -Path "path\to.xml")).Peoples.Person.Name

The above will return all matching nodes as well. It's not as powerful, but it's clean when you know the schema and want one thing out of it quickly.

answered Aug 22, 2017 at 16:43

John Neuhaus

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