Question

Variable $returnedxml is a sql query result forming in xml. I need to get content releasepath \sharespace est1\10.0.1212.00from it

<ReleasePath>\sharespace	est1\10.0.1212.00</ReleasePath>

我的守则如下:

 $xmldoc= new-object xml.xmldocument

 $xmldoc.load($Returnedxml)

 $xmldoc.releasepath

这里是返回的错误警报:

Exception calling "Load" with "1" argument(s): "Could not find file  C:UsersadminSystem.Xml.XmlDocument ."
At D:connecttods3andinvoke.ps1:47 char:14
+  $xmldoc.load <<<< ($Returnedxml)
+ CategoryInfo          : NotSpecified: (:) [], MethodInvocationException
+ FullyQualifiedErrorId : DotNetMethodException

我认为xml.xml文件是一种蚊帐,看来我错了。因此,我可以做些什么?

Answer 1

由于XML数据的处理对于如此众多的管理任务来说是不可或缺的,因此,Powerhell有一个XML型加速器。因此,这也将发挥作用:

[xml]$xmldoc = $returnedxml
$xmldoc.releasepath

Answer 2

我刚刚把它读作一个示意图......

$file = [IO.File]::ReadAllText($filename)

然后利用Xpath从中获取价值......

$releasePath = $file | SelectXml "//ReleasePath"

XPath对于从XML档案中提取物品,比使用xmldoc更简单( wise智)来说,实在是强大的。

Answer 3

页: 1 但是,它目前是<代码>System.Xml.XmlDocument的物体。

So change your variable and then you can read the file.

或者,如果您在<条码>上已经有XmlDocument的标语的话。页: 1 两者都属于同一类别。公正使用费

Answer 4

You are using the wrong load; that one is for files. Use LoadXML instead:

$xmldoc.LoadXml($Returnedxml)

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