Powershell v2 :: Load a COM Interop DLL

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I have this DataLink DLL on my system - Interop.MSDASC.dll I am trying to load the same from Powershell like this - 

[Reflection.Assembly]::LoadFile("C:\Interop.MSDASC.dll") | out-null

But, I get the following error - 

Exception calling "LoadFile" with "1" argument(s): "Could not load file or assembly 'Interop.MSDASC.dll' or one of its dependencies.  is not a 
valid Win32 application. (Exception from HRESULT: 0x800700C1)"
At line:1 char:32
+ [Reflection.Assembly]::LoadFile <<<< ("C:\Interop.MSDASC.dll") | out-null
    + CategoryInfo          : NotSpecified: (:) [], MethodInvocationException
    + FullyQualifiedErrorId : DotNetMethodException

How do I correctly load this ?

powershell
powershell-2.0
asked on Stack Overflow Apr 20, 2012 by Angshuman Agarwal

3 Answers

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This is a 32 bit COM object and therefore you must load it from a 32 bit instance of PowerShell. To do this on a 64 bit version of Windows you can execute powershell.exe or powershell_ISE.exe in this folder: %SYSTEMROOT%\SysWow64\windowspowershell\v1.0

And, this is the complete code:

[Reflection.Assembly]::LoadFile("C:\Interop.MSDASC.dll") 
$dataLinkInstance = new-object MSDASC.DataLinksClass
$dataLinkInstance.WriteStringToStorage("C:\\FrmPowershell.udl", "Provider=SQLOLEDB.1;", 2)
answered on Stack Overflow Apr 23, 2012 by Angshuman Agarwal • edited Jan 23, 2018 by Angshuman Agarwal
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I've just downloaded it from http://datadictionary.codeplex.com/ and load assembly in the same way you use and no issue come:

 [System.Reflection.Assembly]::LoadFile( "c:\Program Files\DataDictionaryCreator\Interop.MSDASC.dll")

GAC    Version        Location
---    -------        --------
False  v2.0.50727     c:\Program Files\DataDictionaryCreator\Interop.MSDASC.dll

Are you maybe on a x64 operative system? if yes read here http://datadictionary.codeplex.com/workitem/28807

answered on Stack Overflow Apr 20, 2012 by CB. • edited Apr 20, 2012 by CB.
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$comInterOp = "C\Temp\Interop.YourAssembly.dll"
[System.Reflection.Assembly]::LoadFile($comInterOp)
$yourClassObj = new-object YourAssembly.YourClassNameClass
$yourResult = $yourClassObj.YourMethod()
answered on Stack Overflow Mar 10, 2020 by Ramakrishna Talla

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