Add element in XML value null

This looks like PowerShell. Here is code for using Xml Linq net library

using assembly System.Xml.Linq

$filename = "c:\temp\test.xml"
$doc = [System.Xml.Linq.XDocument]::Load($filename)
$arrayOfSvc = $doc.Root

$child = [System.Xml.Linq.XElement]::new([System.Xml.Linq.XName]::Get('Svc'))
$arrayOfSvc.Add($child)
$doc

tl;dr

• For an immediate fix, change .Load() to .LoadXml()

• A robust and more efficient approach is to pass the file path directly to the .Load() method:

$drive = "E:"

# Note: Be sure to use a *full path*.
$filePath = "$drive\Alarm\Services\ProgWatch\ProgWatchServices.xml"

$doc = New-Object System.Xml.XmlDocument
$doc.Load($filePath)
$child = $doc.CreateElement("Svc") # An empty new element - see also below.
$doc.DocumentElement.AppendChild($child)

Populating a new XML element before insertion:

The above only creates an empty <Svc> element.
To populate your new element it, you have two options:

• Use methods to iteratively create the desired structure; e.g.:

  $child = $doc.CreateElement('Svc')
  $grandChild1 = $doc.CreateElement('Name')
  $grandChild1.InnerText = 'newName'
  $grandChild2 = $doc.CreateElement('DisplayName')
  $grandChild2.InnerText = 'newDisplayName'
  # ...
  $null = $child.AppendChild($grandChild1)
  $null = $child.AppendChild($grandChild2)
  # ...
  # Finally, insert the completed new child element into the document
  $null = $doc.DocumentElement.AppendChild($child)
  

• More conveniently, use an XmlDocumentFragment instance, obtained via .CreateDocumentFragment(), to construct the new element from an XML string representing the desired structure; e.g.:

  $child = $doc.CreateDocumentFragment()
  $child.InnerXml = '
    <Svc>
      <Name>{0}</Name>
      <Displayname>{1}</Displayname>
      <Activ>{2}</Activ>
      <CommadState>{3}</CommadState>
      <Status>{4}</Status>
    </Svc>
   ' -f 'newName', 'newDisplayName', 'false', 'NoCommand', 'Suspended'
  # Insert the completed new child element into the document
  $null = $doc.DocumentElement.AppendChild($child)
  

Background information re loading XML from files:

Your $filePath variable - despite its name - contains XML text, because you used Get-Content with your file path.

Therefore, the immediate fix is to use the .LoadXml() rather than the .Load() method of the System.Xml.XmlDocument class (which you can refer to as [xml] in PowerShell).

However, this two-step approach to loading XML from a file is best avoided, for the following reasons:

• Efficiency: Using the .Load() method directly with a file path requires only a single call.

  • Note: Be sure to always pass a full path to .NET methods, because .NET's working directory usually differs from PowerShell's.

• Character encoding:

  • The .Load() method defaults to UTF-8 when reading and notably respects an encoding attribute in the XML declaration (e.g., e.g., <?xml version="1.0" encoding="ISO-8859-1"?>) if the character-encoding is a different one.

  • By contrast, Get-Content does not do that (it never interprets the content of a file) and in Windows PowerShell defaults to ANSI encoding, i.e. the encoding implied by the system's active legacy ANSI code page (by contrast, PowerShell (Core) 7+ now commendably default to (BOM-less) UTF-8, across all cmdlets).

  • In other words: Get-Content may misinterpret your XML file.

    • In combination with the [xml] type accelerator, using Get-Content is seductive, but not robust. If you're willing to assume that Get-Content doesn't misinterpret an XML file or you know the actual encoding and specify it via -Encoding, the following shortcut is possible; note the use of -Raw to read the entire file as a whole, which is much faster than the line-by-line reading Get-Content performs by default)

       # Convenient, but NOT ROBUST
       # May need an -Encoding argument.
       $doc = [xml] (Get-Content -Raw $filePath)
      

    • The concise formulation of the robust approach (as shown at the top) is:

       # Still concise, and while more complex, it is ROBUST
       # Note the need for Convert-Path to ensure that a *full path* is passed.
       ($doc = [xml]::new()).Load((Convert-Path $filePath))