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, Jul 11 '13 at 01:19
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, Jul 11 '13 at 01:19

score 36 · Answer 1 · answered Jul 11 '13 at 09:03

36

These two lines should suffice:

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

answered Jul 11 '13 at 09:03

Ansgar Wiechers

193,178
25
254
328

1

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} – Sum1sAdmin Mar 02 '17 at 12:57
If you have a new question: please post a new question. Don't followup on someone else's question in comments. – Ansgar Wiechers Mar 02 '17 at 13:03
sorry, I've posted as a new question http://stackoverflow.com/questions/42561059/powershell-how-to-cast-all-xml-files-as-xml-objects – Sum1sAdmin Mar 02 '17 at 16:43

score 25 · Answer 2 · answered Aug 12 '14 at 07:40

25

How about one line?

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

answered Aug 12 '14 at 07:40

CommonToast

688
12
19

3

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. – John Neuhaus Aug 22 '17 at 16:36

score 16 · Accepted Answer · answered Jul 11 '13 at 01:58

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
}

score 7 · Answer 4 · answered Aug 22 '17 at 16:43

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.

Selecting attributes in xml using xpath in powershell

4 Answers4

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