如何通过powershell从xml文件中获取一些内容？
变量$returnedxml是一个以xml形式形成的sql查询结果。我需要从中获取内容'releasepath'\\sharespace\test1\\10.0.1212.00这
<ReleasePath>\\sharespace\test1\\10.0.1212.00</ReleasePath>
是我的代码：
$xmldoc= new-object xml.xmldocument
$xmldoc.load($Returnedxml)
$xmldoc.releasepath
这是返回的错误警报：
Exception calling "Load" with "1" argument(s): "Could not find file 'C:\Users\admin\System.Xml.XmlDocument'."
At D:\connecttods3andinvoke.ps1:47 char:14
+ $xmldoc.load <<<< ($Returnedxml)
+ CategoryInfo : NotSpecified: (:) [], MethodInvocationException
+ FullyQualifiedErrorId : DotNetMethodException
我认为xml.xmldocument是一个.net类，看来我错了。那我能做什么呢？
Variable $returnedxml is a sql query result forming in xml. I need to get content 'releasepath'\\sharespace\test1\\10.0.1212.00from it
<ReleasePath>\\sharespace\test1\\10.0.1212.00</ReleasePath>
Here are my code:
$xmldoc= new-object xml.xmldocument
$xmldoc.load($Returnedxml)
$xmldoc.releasepath
Here are the returned error alarm:
Exception calling "Load" with "1" argument(s): "Could not find file 'C:\Users\admin\System.Xml.XmlDocument'."
At D:\connecttods3andinvoke.ps1:47 char:14
+ $xmldoc.load <<<< ($Returnedxml)
+ CategoryInfo : NotSpecified: (:) [], MethodInvocationException
+ FullyQualifiedErrorId : DotNetMethodException
I thought xml.xmldocument is a .net class, seems that I was wrong. So what can I do then?
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由于 XML 数据的处理对于许多管理任务来说都是不可或缺的，因此 Powershell 有一个 XML 类型加速器。所以这也可以工作：
Since the handling of XML data is so integral to so many management tasks, Powershell has an XML Type Accelerator. So this would work as well:
我只是将其读入字符串...
然后使用 xpath 从中获取值...
XPath 对于从 XML 文件中提取内容非常强大，比使用 xmldoc 简单得多（编码方面）
I just read it into a string ...
then use xpath to get values out of it ...
XPath is really powerful for pulling things out of an XML file, much simpler (coding wise) than using xmldoc
您的变量
$Returnedxml必须是带有绝对路径的文件名。但目前它是 System.Xml.XmlDocument 类的对象。因此，更改您的变量，然后您就可以读取该文件。
或者，另一方面，如果您在
$Returnedxml中已经有一个 XmlDocument 对象，那么您不必将其读入$xmldoc中。两人都来自同一个班级。只需使用 $ReturnedxmlYour variable
$Returnedxmlmust be a file name with absolute path. But currently it is an object of classSystem.Xml.XmlDocument.So change your variable and then you can read the file.
Or on the other hand if you already have an object of XmlDocument in
$Returnedxmlthen you do not have to read it into$xmldoc. Both are from the same class. Just use $Returnedxml您使用了错误的负载；那个是用于文件的。使用
LoadXML代替：You are using the wrong load; that one is for files. Use
LoadXMLinstead: