Solved

Extract Just Path Location from File Picker  vba Access

Posted on 2016-08-17
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Last Modified: 2016-08-17
vba

Using

' To extract only the filename from the path, you can do the following:
   varFilename = Mid(fDialog.SelectedItems(1), InStrRev(fDialog.SelectedItems(1), "\") + 1, Len(fDialog.SelectedItems(1)))
  

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The Above gives me
myexcelfile.xls
from
C:\Program Files\MyData\myexcelfile.xls



What I need:
I need to extract just the file Path Location :
C:\Program Files\MyData\

This will not work: it gives me the whole file location path and name:  C:\Program Files\MyData\myexcelfile.xls



thanks
fordraiders
0
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Question by:fordraiders
3 Comments
 
LVL 47

Accepted Solution

by:
Dale Fye (Access MVP) earned 500 total points
ID: 41759835
try:

varPath = LEFT(fDialog.SelectedItems(1), InStrRev(fDialog.SelectedItems(1), "\"))
0
 
LVL 26

Expert Comment

by:Shaun Kline
ID: 41759838
Try this:
   varPath = Left(fDialog.SelectedItems(1), Len(fDialog.SelectedItems(1)) - InStrRev(fDialog.SelectedItems(1), "\") - 1, )
0
 
LVL 3

Author Closing Comment

by:fordraiders
ID: 41759941
Thanks dale.
0

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