Getting Path Information From A Text String

saltkev

Active Member
Joined
Oct 21, 2010
Messages
324
Office Version
  1. 2013
Platform
  1. Windows
Good Morning

I hope someone can help. I have a file path as shown below. I need to remove the file name from the end leaving just the path as shown in the second line. The path and file names will change.


C:\Users\keames1\Desktop\2 - Current DD + N R9\WCT Station 130 DDR9.xlsm
[TABLE="width: 70"]
<tbody>[TR]
[TD][TABLE="width: 500"]
<tbody>[TR]
[TD]C:\Users\keames1\Desktop\2 - Current DD + N R9\[/TD]
[/TR]
[TR]
[TD][/TD]
[/TR]
[TR]
[TD][/TD]
[/TR]
</tbody>[/TABLE]
[/TD]
[/TR]
</tbody><colgroup><col></colgroup>[/TABLE]
Many Thanks
[TABLE="width: 70"]
<tbody>[TR]
[TD][/TD]
[/TR]
</tbody><colgroup><col></colgroup>[/TABLE]
 

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Hi, here is one option you can try:


Excel 2013/2016
AB
1C:\Users\keames1\Desktop\2 - Current DD + N R9\WCT Station 130 DDR9.xlsmC:\Users\keames1\Desktop\2 - Current DD + N R9\
Sheet1
Cell Formulas
RangeFormula
B1=TRIM(LEFT(SUBSTITUTE(A1,"",REPT(" ",255),LEN(A1)-LEN(SUBSTITUTE(A1,"",""))),255))&""
 
Upvote 0
Hi FormR, Why the &"" at the end?

Hi Pgc - there should be a backslash within the double quotes - the forum must have parsed it thinking it was HTML.

=TRIM(LEFT(SUBSTITUTE(A1,"\",REPT(" ",255),LEN(A1)-LEN(SUBSTITUTE(A1,"\",""))),255))&"\"

Thanks for spotting.

EDIT: it got parsed in a few places - updates in red.
 
Last edited:
Upvote 0
You are right, I had not examined the formula just saw the &"" at the end.

But now I looked at the formula and you miss another \ in the Substitute. Must be the same problem as the other.

EDIT: I see that you already spotted it.
 
Last edited:
Upvote 0
Thanks For The Input Guys. However, I was looking for a VBA Solution, I already have a cell based solution.

Again Many Thanks

Kev
 
Upvote 0
I was looking for a VBA Solution

Hi, that's always worth mentioning at the outset.

One option, for example:

Code:
Dim S As String
S = "C:\Users\keames1\Desktop\2 - Current DD + N R9\WCT Station 130 DDR9.xlsm"
MsgBox Left(S, InStrRev(S, "\"))
 
Upvote 0

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