按属性值选择 XML 元素并添加元素

问题描述

我有这个结构的 XML 文件:

I have this XML-file with this structure:

<?xml version="1.0" encoding="utf-8"?>
<company>
<category>
    <category1 name="Office1">
        <category2 name="Project1">
            <category3 name="Test1"/>
            <category3 name="Test2"/>
        </category2>
        <category2 name="Project2">
            <category3 name="Test1"/>
            <category3 name="Test2"/>
            <category3 name="Test3"/>
        </category2>
     </category1>

     <category1 name="Office2">
        <category2 name="Project1">
            <category3 name="Test1"/>
            <category3 name="Test2"/>
        </category2>
        <category2 name="Project2">
            <category3 name="Test1"/>
            <category3 name="Test2"/>
            <category3 name="Test3"/>
        </category2>
      </category1>
</category>
</company>

我想添加一行到公司 -> 类别 -> 类别 1 "Office2" -> 类别 2 "项目 2"该行是:

I want to add a line to company -> category -> category1 "Office2" -> category2 "Project2"The line is:

<category3 name="Test4"/>

我已经试过了:

$Path = "C:\file.xml"
$xml = [xml](get-content $Path)
$xml.Load($Path)
$test = $xml.company.category
$test.category1 *what to do here*

我知道如何用一个子元素来做到这一点，以及如何克隆和添加.但我不知道从哪里开始.

I know how to do this with one sub-element, and how to clone and add. But I don't know where to start with this one.

推荐答案

不知道是否有更短的方法，但这应该有效:

Don't know if there is a shorter way, but this should work:

$Path = "C:\file.xml"
$xml = [xml](get-content $Path)
$xml.Load($Path)
$target = (($xml.company.category.category1|where {$_.name -eq "Office2"}).category2|where {$_.name -eq "Project2"})
$addElem = $xml.CreateElement("Category3")
$addAtt = $xml.CreateAttribute("name")
$addAtt.Value = "Test4"
$addElem.Attributes.Append($addAtt)
$target.AppendChild($addElem)
$xml.Save("C:\file1.xml")

这里的要点是使用 where 获取具有给定属性值的元素以及创建新元素和新属性.

The main points here are the usage of where to get the elements with the given attribute values and the creation of a new element and a new attribute.

获取目标"元素的另一种可能解决方案是使用 XPath:

Another possible solution to get the "target" element is the usage of XPath:

$target = $xml.SelectSingleNode('//company/category/category1[@name="Office2"]/category2[@name="Project2"]')

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