Powershell:从 xml 文件中读取值 [英] Powershell: Read value from xml files
问题描述
我需要一些有关 PowerShell 的帮助,请.应该很简单:
I need some with help with PowerShell, please. It should be pretty easy:
我有一个子目录列表,每个子目录中都有一个 xml 文件.我想打开每个 xml 文件并打印一个节点的值.节点始终相同,因为 xml 文件实际上是来自 Visual Studio 的项目文件 (*.csproj).
I have a list of subdirectories, with a xml file in each one. I want to open each xml file and print the value of one node. The node is always the same, as the xml files are actually project files (*.csproj) from Visual Studio.
我已经得到了文件列表:get-item ** \ *.csproj
I already got the list of files: get-item ** \ *.csproj
我该如何进行?
推荐答案
要获取 csproj 文件,请使用 Get-ChildItem
To get the csproj files use Get-ChildItem
Get-ChildItem c:\myProjects *.csproj -recurse
然后你可以使用例如Select-Xml
像这样:
Then you can use e.g. Select-Xml
like this:
$ns = @{ defaultNamespace = "http://schemas.microsoft.com/developer/msbuild/2003" }
Get-ChildItem c:\myProjects *.csproj -recurse |
Select-Xml -xpath '//defaultNamespace:PropertyGroup[1]' -namespace $ns |
Select-Object -expand Node
您必须更正默认命名空间.我现在手头没有 csproj 文件.
(有关 PowerShell 中 xml 和 xpath 的更多信息,请参阅 http://huddledmasses.org/xpath-and-namespaces-in-powershell/)
You have to correct the default namespace. I have no csproj file by hand right now.
(for more info about xml and xpath in PowerShell see http://huddledmasses.org/xpath-and-namespaces-in-powershell/)
需要Select-Object
来扩展实际的节点属性.
The Select-Object
is needed to expand the actual node property.
如果你想像处理对象一样处理 xml,你可以使用这个:
If you would like to work with xml like with object, you can use this:
Get-ChildItem c:\myProjects *.csproj -recurse |
% { $content = [xml](gc $_.FullName); $content.Project.PropertyGroup[someindex...] }
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