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How to open up Excel Workbook from a button click selection

Posted on 2011-09-30
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Last Modified: 2012-08-14
I have posted code below that will open up the Excel Spreadsheet however I can;t seem to figure out how to open up an individual workbook for selection.

Thanks
Private Sub btnXYZFile_Open_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles btnXYZFile_Open.Click
        Dim XlFilePath As String
        Me.dlgOpenPartfile.InitialDirectory = _Default_Definition_FileOpen_Dir
        If Me.dlgOpenPartfile.ShowDialog(Me) = Windows.Forms.DialogResult.OK Then
            '  Opens the target Excel File
            Dim ExcelObj As New Microsoft.Office.Interop.Excel.Application
            ExcelObj.Visible = True

        End If
    End Sub

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Question by:cmdolcet
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6 Comments
 
LVL 83

Expert Comment

by:CodeCruiser
ID: 36891844
Not sure what you mean. Do you want to open an Excel file in Excel? Use Process.Start("path of xls file")
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Author Comment

by:cmdolcet
ID: 36891862
how do I get that path. I can open up a selection window in the code above the Excel Object but when I select the File in the folder the workbook doesn;t open
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LVL 83

Expert Comment

by:CodeCruiser
ID: 36892001
The path will be Me.dlgOpenPartfile.FileName property.
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Author Comment

by:cmdolcet
ID: 36892257
How can I capture the value selected ?
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LVL 40

Accepted Solution

by:
Jacques Bourgeois (James Burger) earned 500 total points
ID: 36892319
Code Cruiser told you how to get the value. You can use it this way:

ExcelObj.WorkBooks.Open(dlgOpenPartfile.FileName)
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Author Closing Comment

by:cmdolcet
ID: 36892330
opps I didn't read that carefully
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