Macro open file and go back

I have a macro that will open a dialog box to allow the user to open another file. There are a couple of things that I need help with. First, I would like the code to open another excel file to run as soon as the user opens this initial file, let's call it Workbook1. Then after the user selects the file that will provide input into Workbook1, let's call it Workbook2, I want Workbook1 to be the file that the user sees again. From there, the user will either select other buttons to run other VB, or other activities. My open file code is below:

' Open a file

Sub Open_File()
Dim sFil As String
Dim sTitle As String
Dim sWb As String
Dim iFilterIndex As Integer
On Error GoTo err_handler
' Set up list of file filters
sFil = "Excel Files (*.xls),*.xls"
' Display *.xls by default
iFilterIndex = 1
' Set the dialog box caption
sTitle = "Select File to Zip"
' Get the filename
sWb = Application.GetOpenFilename(sFil, iFilterIndex, sTitle)

Workbooks.Open Filename:=sWb
Exit Sub
err_handler:
MsgBox "No selection made"

End Sub

Any assistance with this is much appreciated. Thanks in advance.

Thanks for the code. When I run it, i am able to select a file, but it does not open and jumps right to "No selection made".

Any thoughts?

Ok. First chance to try this again. Commented out the line and now I get a "Type Mismatch" message when the code steps to

If sWb <> False Then

What would cause this and how to repair it?

Thanks.

Excellent! This worked.

Thank you.

Another question on this. So I now have 2 files open, a main file and this other file that the user has selected. I need to take certain bits of information from the file that was selected and copy into the main file. How would I assign the filename as a variable so that I can jump back and forth between the main file and the file that was selected? For example, I need to look at Column C of the selected file, find the first non-blank cell, copy the contents, then move back to the main file and past that information into a specific cell.

Thanks again!

Thanks for the response. I put this into the code and changed to the cell references needed for the test. I then stepped through the code. I am able to open a file, but when I get to the code above, I get an error:

Run-time error '438': Object doesn't support this property or method


Sub Update_Feedback()
'

Dim WBSrc As Workbook
Dim WBDest As Workbook
Dim sFil As String
Dim sTitle As String
Dim sWb As String
Dim iFilterIndex As Integer
' On Error GoTo err_handler
Set WBSrc = ThisWorkbook
' Set up list of file filters
sFil = "Excel Files (*.xl*),*.xl*"
' Display *.xls by default
iFilterIndex = 1
' Set the dialog box caption
sTitle = "Select Auditor Feedback File to Open"
' Get the filename
sWb = Application.GetOpenFilename(sFil, iFilterIndex, sTitle)

If sWb <> "False" Then
Application.ScreenUpdating = False
Set WBDest = Workbooks.Open(Filename:=sWb)
WBSrc.Activate
Application.ScreenUpdating = True

WBDest.Range("F382") = WBSrc.Range("E9")
WBDest.Range("G382") = WBSrc.Range("F9")
WBDest.Range("H382") = WBSrc.Range("G9")
WBDest.Range("I382") = WBSrc.Range("H8")

End If
Exit Sub
err_handler:
MsgBox "No selection made"

End Sub

Thank you. I have updated the code as noted. I have ensured that the worksheet names match between the workbook and the code. I have changed the worksheet names to ensure no spaces as well. I now get an error of "Subscript out of Range".

Sorry for all the questions.

Thanks for your help.

Destination.xlsmSource.xlsx

Please see the attached. When I run this code, instead of the Destination getting information from the Source, it goes the other way. Take a look at Source.xlsx before you run the code. There are no blank cells. When you run the code from Destination.xlsm, you will see a blank cell in Source.xlsx. The need is to take the information from Source and paste into Destination.

Thanks.

EXCELLENT! Thank you very much. Have a great weekend.