How to get the node value from xml using PowerShell script?

Question

<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE suite SYSTEM "http://testng.org/testng-1.0.dtd">
<suite name="MultipleSuite">
<suite-files>
    <suite-file path="samplexml_1.xml"></suite-file>
</suite-files>
</suite>
<!-- Suite -->

here how to get the 'samplexml_1.xml' name ?

Tomalak · Accepted Answer · 2018-09-03 12:10:58Z

1

Use XPath and the Select-Xml cmdlet:

$suite_file = Select-Xml "//suite-file" yourfile.xml

This selects an array of <suite-file> nodes (in your example, there is only one match). You can access the result like this:

$suite_file.Node.path

prints

samplexml_1.xml

answered Sep 3, 2018 at 12:10

Tomalak

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Kirill Pashkov · Accepted Answer · 2018-09-03 11:50:18Z

0

Get content of xml file to [xml] type variable. Like so:

[xml]$xml = Get-Content file.xml
$xml.suite.'suite-files'.'suite-file'.path
#samplexml_1.xml

answered Sep 3, 2018 at 11:50

Kirill Pashkov

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1 Comment

Tomalak Over a year ago

Don't do [xml]$xml = Get-Content file.xml. This will break the data when the file is not in the encoding Get-Content expects. $xml = New-Object xml; $xml.Load("file.xml") is much better, because this uses the encoding detection of the XML parser.

TobyU · Accepted Answer · 2018-09-03 11:51:08Z

0

This would work but is probably not the best solution depending on what you want to achieve at the end.

$xml = [xml]'<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE suite SYSTEM "http://testng.org/testng-1.0.dtd">
<suite name="MultipleSuite">
<suite-files>
    <suite-file path="samplexml_1.xml"></suite-file>
</suite-files>
</suite>'

$xml.suite."suite-files"."suite-file".path

answered Sep 3, 2018 at 11:51

TobyU

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