Opening a Workbook with file path reference in a cell

[NextWorldAngel]

I am Trying to Open a Workbook with the file name referenced in a cell. Below is what I am using but it is not working can anyone tell me what I need to change?

 Workbooks.Open Filename:=Sheet("Support Info").Range("F31")

 

[wigi]

Workbooks.Open(Sheet("Support Info").Range("F31"))

or

Workbooks.Open([Support Info!F31])

 

[Trevor G]

I am Trying to Open a Workbook with the file name referenced in a cell. Below is what I am using but it is not working can anyone tell me what I need to change?

 Workbooks.Open Filename:=Sheet("Support Info").Range("F31")

Question does the cell also contain the file extension of .xls?

Would it be useful to show the content of the cell in your thread to help get a solution.

 

[NextWorldAngel]

That didn't work here is my the path that I have in the cell:

C:\Users\NextWorldAngel\Desktop\USPS.csv

Thanks,

 

[wigi]

Wit a simple CSV file and the path written in cell A1, this works perfectly for me:

Sub tst()
   Workbooks.Open (Sheets("Sheet1").Range("A1"))
End Sub

Note that this is completely analogous to what I provided you earlier.