How to check if a file is corrupt?-CodePudding

I have a loop that opens word documents however there are a few that are corrupted, I can't open them manually either,

The code I'm using to open is,

Set newdoc = objWord.Documents.Open(saveFolder & Filename)

This fails with the error

"Run-time error 5792, the file appears to be corrupted".

Is it possible to tell that the file is corrupt before trying to open it?

CodePudding user response：

No, but you can test it while trying to open it.

On Error Resume Next ' deactivate error reporting
Set newdoc = Nothing
Set newdoc = objWord.Documents.Open(saveFolder & Filename)

' check if trying to open it produced an error
If Err.Number = 5792 Then
    Msgbox "'" & saveFolder & Filename & "' is corrupt."
    Err.Clear
End If

On Error Goto 0  ' re-activate error reporting. Do not forget this!

' if an error occured during opening then newdoc is nothing
If Not newdoc Is Nothing

    ' do what you want to do with the opened file

Else

    ' do what you want to do with a corrupt file

End If

CodePudding user response：

your best bet here is to use error handling.

Something like

On Error GoTo skip
Set newdoc = objWord.Documents.Open(saveFolder & Filename)
On Error GoTo 0
Exit Function
skip: