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 ?

Answer 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

Answer 2

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

Answer 3

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