How to open DWG file in Autocad 2009 in C# ?

[wperaud]

Hello,

I'm developping a dll for Autocad 2009.

I would like to know, how to open a DWG file ?

I tried :
Database db = new Database();
db.ReadDwgFile(file.Path, FileOpenMode.OpenForReadAndAllShare, true, "");
---> does nothing

Autodesk.AutoCAD.ApplicationServices.Application.DocumentManager.Open(file.Path);
--> Error concerning Execution Context.

Thank you

[wperaud]

Oh !! So obvious !! I don't know why I didn't think about it ^^

Thank you

[ViktorCAD]

Tony, help me understand your answer (its been several months since i did anything to autocad .net api)...

I'm seeing this problem in my project, i have a form with a list box (non-wpf) and in the double click event of the listbox i wired it up to go to a function that opens a file. I do something like this:

docs = Application.DocumentManager
docs.Open(fname, False, "")

Of course like you mention this opens the file but it is not active, and I understand the reason because the main command that executes my form is not run out of a session but out of a document.

So i'll make another command, that runs in session and have it open a document, right? but how do i pass it the filename? i've never done that...

so in my command:

_
Public Sub OpenDocumentCmd()


End Sub

Can i just pass to the sub?

Thanks,
Viktor.

[Anonymous]

First, if your dialog box is being shown modally, nothing is going to happen
until it's closed, so do you really want to use a modal dialog?

The command you invoke from your control event handler via
SendStringToExeucte() can access whatever data it needs, from static members
whose values you set just before you invoke the command.

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wrote in message news:6204854@discussion.autodesk.com...
Tony, help me understand your answer (its been several months since i did
anything to autocad .net api)... I'm seeing this problem in my project, i
have a form with a list box (non-wpf) and in the double click event of the
listbox i wired it up to go to a function that opens a file. I do something
like this: docs = Application.DocumentManager docs.Open(fname, False, "") Of
course like you mention this opens the file but it is not active, and I
understand the reason because the main command that executes my form is not
run out of a session but out of a document. So i'll make another command,
that runs in session and have it open a document, right? but how do i pass
it the filename? i've never done that... so in my command: _ Public Sub
OpenDocumentCmd() End Sub Can i just pass to the sub? Thanks, Viktor.

[wperaud]

Here is an example :
{code}
public class FileManager{

// Defines the current file managed by the application
public static String CurrentFilePath;

// Set an open the current file
public void OpenFile(String filePath)
{
// Set
this.CurrentFilePath = filePath;
// Call the Autocad custom command in order to open the file
acadApp.DocumentManager.MdiActiveDocument.SendStringToExecute("OpenFile\n", false, false, false);
}

}
{code}

{code}
[CommandMethod("OpenFile", CommandFlags.Session)]
static public void OpenFile()
{
DocumentCollection dm = acadApp.DocumentManager;
if (System.IO.File.Exists(FileManager.CurrentFilePath)
dm.Open(FileManager.CurrentFilePath, false);
}
{code}

Instead of a static attribute or method, the path can be retrieved thanks to a method.
As an example, I use the Singleton Pattern to the file manager, and get the filepath like this :
{code}
FileManager.Instance().GetCurrentFilePath();
{code}