电源外壳：
$doc = new-object System.Xml.XmlDocument
$doc.Load($filename)
$items = Select-Xml -Xml $doc -XPath '//item'
$items | foreach {
$item = $_
write-host $item.name
}
我没有输出
XML：
<?xml version="1.0" encoding="UTF-8"?>
<submission version="2.0" type="TREE" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:noNamespaceSchemaLocation="TREE.xsd" xmlns="some/kind/of/tree/v1">
<group>
<item></item>
<item></item>
<item></item>
</group>
<submission>
你有一些问题在继续.首先,您需要在XPath模式中指定命名空间,XML格式不正确(结束标记不是结束标记),而Select-Xml直接返回XmlInfo而不是XmlElement.试试这个：
$xml = [xml]@'
<submission version="2.0" type="TREE" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:noNamespaceSchemaLocation="TREE.xsd" xmlns="some/kind/of/tree/v1">
<group>
<item></item>
<item></item>
<item></item>
</group>
</submission>
'@
$ns = @{dns="some/kind/of/tree/v1"}
$items = Select-Xml -Xml $xml -XPath '//dns:item' -Namespace $ns
$items | Foreach {$_.Node.Name}
