按属性值选择 XML 元素并添加元素
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Select XML element by attribute value and add an element
提问by alexfyren
I have this XML-file with this structure:
我有这个结构的 XML 文件：
<?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>
I want to add a line to company -> category -> category1 "Office2" -> category2 "Project2" The line is:
我想在 company -> category -> category1 "Office2" -> category2 "Project2" 中添加一行，该行是：
<category3 name="Test4"/>
I've tried this:
我试过这个：
$Path = "C:\file.xml"
$xml = [xml](get-content $Path)
$xml.Load($Path)
$test = $xml.company.category
$test.category1 *what to do here*
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.
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。
回答by Ocaso Protal
Don't know if there is a shorter way, but this should work:
不知道是否有更短的方法，但这应该有效：
$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")
The main points here are the usage of whereto get the elements with the given attribute values and the creation of a new element and a new attribute.
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。
Another possible solution to get the "target" element is the usage of XPath:
获取“目标”元素的另一种可能解决方案是使用 XPath：
$target = $xml.SelectSingleNode('//company/category/category1[@name="Office2"]/category2[@name="Project2"]')
