How to get some content from xml file by powershell?

Question

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?

EBGreen · Accepted Answer · 2011-08-03 13:14:17Z

4

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:

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

answered Aug 3, 2011 at 13:14

EBGreen

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2 Comments

JasonMArcher Over a year ago

According to the error message, it already is an XML document. So you can skip the first line.

JasonMArcher Over a year ago

But look at the file it is trying to find, 'C:\Users\admin\System.Xml.XmlDocument'. You can see there is a type name in the string, which is the default ToString() operation. That tells us that $Returnedxml is an XML document, because when converted to a string, you get the type name. Which is then treated as a relative path, so you get the result 'C:\Users\admin\System.Xml.XmlDocument'.

Antony Scott · Accepted Answer · 2011-08-03 15:35:22Z

2

I just read it into a string ...

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

then use xpath to get values out of it ...

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

XPath is really powerful for pulling things out of an XML file, much simpler (coding wise) than using xmldoc

answered Aug 3, 2011 at 15:35

Antony Scott

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Comments

mailq · Accepted Answer · 2011-08-03 11:30:33Z

1

Your variable $Returnedxml must be a file name with absolute path. But currently it is an object of class System.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 $Returnedxml then you do not have to read it into $xmldoc. Both are from the same class. Just use $Returnedxml

answered Aug 3, 2011 at 11:30

mailq

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Matt · Accepted Answer · 2013-03-04 15:09:23Z

0

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

$xmldoc.LoadXml($Returnedxml)

edited Mar 4, 2013 at 15:09

user142162

answered Aug 4, 2011 at 4:49

Matt

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