How To Get Function OpenFileDialog To Return Path

Mar 15, 2011

I am using the open file dialog box. Right now the function looks like

Private Function OpenFileDialog1_FileOK(...)
File_Path_Name = OpenFileDialog1.Filename.ToString()
End Function

The thing is it isn't clear when this is called. I think it is called when I go through
OpenFileDialog1.ShowDialog()

I want to return a value that is the Filename path to a variable like
file_path = Function()
and not use a global variable. But it doesn't seem like this is possible. Do you know how to get the function OpenFileDialog to return the file path?

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[Code]...

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[Code]...

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Here is my snippet:

private void btnBrowseCInv_Click(object sender, EventArgs e)
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[code]....

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2.Exit

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[Code]....

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[Code]...

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OpenFileDialog - Getting Specific File Type From My Computer Using The .getfiles Function

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I'm doing a project now. I'm having a problem on getting specific file type from my computer using the .getfiles function.

Here's my codes.

Private Sub dlgOpen_FileOk(ByVal sender As System.Object, ByVal e As System.ComponentModel.CancelEventArgs) Handles dlgOpen.FileOk

Dim filepath as String = "C:"
dlgOpen.CheckFileExists = True

[CODE]...

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[code]...

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Public Class LoadSet
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[code]....

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How would I do that? Sample code is below that demonstrated that the file I opened contained the full path and file name.

I want to extract just the path and serialize that to the user.config file as a UserSetting value. Then next time the user opens the dialog box, it uses that saved path string to go immediately to the location previously used.

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[Code].....

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vb.net
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[Code] .....

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[Code]....

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[Code]....

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[code].....

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[Code].....

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[Code].....

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[Code]...

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[Code].....

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