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How to get file path of opened document from OCX object.

I am using OCX control in our program and I need to save links to opened documents, that's why I need to get full path (or just the ServerName) of opened document in OCX control.
If I open some document:
you can find the opened string in Me.AxQlikOCX.DocName. This is OK.

But if you try to open only  “qvp://”  string
a "Connect to Server" window pops-up. Here you can type your server name and choose from available documents. After clicking on "Open" button, document is opened correctly but then when I look at Me.AsQlikOCX.DocName property, there is just "qvp://". (What I was opening.)
Is there any way how to get full file path or just the server name?
I was looking in AxQlikOCX.ActiveDocument object, but I just found the FileName property (which consists of "MyDir/MyFile.qvw") and MyWorkingDirectory property (which probably holds exact path on server "C:\Dir1\Dir2\MyDir"), but I am not able to find something like "qvp://ServerName/MyDir/MyFile.qvw" or just "qvp://ServerName/" which I can concatenate with AxQlikOCX.ActiveDocument.FileName.
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Re: How to get file path of opened document from OCX object.


I am facing same problem,, do you find solution for this??

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