Question

我有一个像这样的xml文件：

<?xml version="1.0" encoding="UTF-8" standalone="yes" ?>
<Item>
<Field Name="Filename">x1</Field>
<Field Name="Year">y1</Field>
<Field Name="Name">z1</Field>
</Item>
<Item>
<Field Name="Filename">xn</Field>
<Field Name="Year">yn</Field>
<Field Name="Name">zn</Field>
</Item>

对于每个Item，使用PowerShell，我想创建一个只包含相关信息的xml文件。我该怎么办？

Answer 1

您可以使用<Item>获取所有SelectNodes()个节点，然后从每个节点中选择Filename字段SelectSingleNode()：

$xml = [xml]@'
<?xml version="1.0" encoding="UTF-8" standalone="yes" ?>
<Items>
<Item>
<Field Name="Filename">x1</Field>
<Field Name="Year">y1</Field>
<Field Name="Name">z1</Field>
</Item>
<Item>
<Field Name="Filename">xn</Field>
<Field Name="Year">yn</Field>
<Field Name="Name">zn</Field>
</Item>
</Items>
'@

foreach($item in $xml.SelectNodes('//Item'))
{
    $Filename = $item.SelectSingleNode('//Field[@Name = "Filename"]').InnerText

    # create your new xml document here, import $Item and save to $Filename
}

Answer 2

您可以尝试这样的事情：

$xml = [xml]@'
<?xml version="1.0" encoding="UTF-8" standalone="yes" ?>
<Items>
<Item>
<Field Name="Filename">x1</Field>
<Field Name="Year">y1</Field>
<Field Name="Name">z1</Field>
</Item>
<Item>
<Field Name="Filename">xn</Field>
<Field Name="Year">yn</Field>
<Field Name="Name">zn</Field>
</Item>
</Items>
'@

$xml.Items.Item | ForEach-Object {
    $filename = $_.SelectSingleNode("Field[@Name='Filename']").innertext
    $file = New-Object xml
    $file.InsertBefore($file.CreateXmlDeclaration("1.0","UTF-8","yes"),$file.DocumentElement) | Out-Null
    $file.AppendChild($file.ImportNode($_,$true)) | Out-Null
    $file.Save("C:\users\frode\Desktop\$($filename).xml")
}

创建子xml文件？

2 个答案: