[Alkis]

Dim excel As New Microsoft.Office.Interop.Excel.Application()
Dim wb As Microsoft.Office.Interop.Excel.Workbook = excel.Workbooks.Open(txtFileLocations.Lines(w))
Dim ws As Microsoft.Office.Interop.Excel.Worksheet = TryCast(excel.ActiveSheet, Microsoft.Office.Interop.Excel.Worksheet)

when the program tries to opens an excel file, it always throws me an error "Office has detected a problem with this file. Editing may hard your computer" and my program crushes.......
The problem appears on the SECOND line, on DIM wb...blahblah..

i went to the office and unclicked the "protected view" checkboxes.. but it still throws me those errors.
my guess is that the office i have in the computer is not the same as the one that my program uses to open excel files (i mean, i loaded/imported some excel.dll files in the program so i guess that will be the one)

But how do i make the program not care if the excel file is "potentially dangerous"? i KNOW it is not.
and i dont understand how did that happen just NOW, out of the blue :S

[Alkis]

people? anyone?

[fixo]

Not sure about if this helps,
anyway take a look at:
http://support.micro...b;en-us;Q320369
http://msdn.microsof...office.11).aspx

[Alkis]

no, unfortunatelly this isnt the case here

[Alkis]

So... can anyone help me a bit with this?

[fixo]

See here, please:

http://www.sevenforu...1968-post1.html

[Alkis]

fixo, on 30 July 2011 - 01:56 AM, said:

See here, please:

http://www.sevenforu...1968-post1.html

unfortunately this doesnt help. if i want to open the file on my computer, i can.

what i can not do, is open the file via my program...

[fixo]

Just tried your code 10 times on my end,
it's working good for me
(VB.NET 2010, Win7, Office 2007)

Imports System.IO
'require reference to Microsoft Excel 12.0 Object Library
Module Module1

    Sub Main()
        If Not File.Exists(txtFileLocations.Lines(w)) Then
            MsgBox("Excel file " & txtFileLocations.Lines(w) & " does not exist")
            Return
        End If
        
        Dim excel As New Microsoft.Office.Interop.Excel.Application()
        Try
            excel.Visible = True
            Dim wb As Microsoft.Office.Interop.Excel.Workbook = CType(excel.Workbooks.Open(txtFileLocations.Lines(w)), Microsoft.Office.Interop.Excel.Workbook)
            Dim ws As Microsoft.Office.Interop.Excel.Worksheet = TryCast(excel.ActiveSheet, Microsoft.Office.Interop.Excel.Worksheet)
        Catch ex As System.Exception
            MsgBox(ex.Message & vbLf & ex.StackTrace)
        End Try
    End Sub

End Module

[Alkis]

It works great when the excel document does not open in protected view

But if it does, then the program cannot open it

[liviuz]

Did you have a resolution to this error as I am getting a similar error myself.

Alkis, on 10 August 2011 - 08:39 AM, said:

It works great when the excel document does not open in protected view

But if it does, then the program cannot open it

[Alkis]

unfortunately i have yet to find a solution

[Alkis]

Anyone? anyone at all???
i really need to be able to open all excel files, protected-view and non-protected view alike

btw, the file that i try to open and makes my life miserable is this:

http://www.opap.gr/documents/10157/30233/kino_2011_04.xls

[fixo]

You have to turn off protected view manually
before of the opening file, I think

Take a look at :
http://www.webtlk.co...in-office-2010/

[liviuz]

Alkis, on 03 September 2011 - 12:53 AM, said:

Anyone? anyone at all???
i really need to be able to open all excel files, protected-view and non-protected view alike

btw, the file that i try to open and makes my life miserable is this:

http://www.opap.gr/documents/10157/30233/kino_2011_04.xls

I managed to solve this problem by using the ProtectedViewWindows object, new in Office 2010. Here is my code, although in C#:

Excel.Application xcl = new Excel.ApplicationClass();
Excel.ProtectedViewWindow pvw = xcl.ProtectedViewWindows.Open(filename);
Excel.Workbook wkBook = pvw.Workbook;
..........

I hope this helps.

[Pit Ufo]

Why don´t you just open the file with the Proccess Class?

Process.start("My File Path")