10

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>
4

1 回答 1

13

你遇到了一些问题。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)并且 Select-Xml 返回 XmlInfo 而不是直接返回 XmlElement。尝试这个:

$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}
于 2013-10-11T03:05:24.390 回答