On Fri, 15 Feb 2008 21:36:13 -0500, Ron Rosenfeld
wrote:

On Fri, 15 Feb 2008 18:07:05 -0800, John Gregory
wrote:

The text strings are random length - file path names:

c:\sampledir1\filename - description

In this example, I want to get the string "filename", but I have many
directories, all with differnt length names.

c:\samplelongname2\filename345 - longer description

etc.

In all cases I want to get the text between the "/" and the "-". The right
and left functions do not work because the number of characters varies.

Any ideas


In your example, the - is surrounded by <space on both sides.

If this is the case in your strings, it would be more robust to look for that
sequence, than just for the "-".

With your string in A1, here is a formula that will extract the string that is
between the last "\" and the last "-" :

=TRIM(MID(A1,1+FIND(CHAR(1),SUBSTITUTE(A1,"\",CHA R(1),
LEN(A1)-LEN(SUBSTITUTE(A1,"\","")))),-1+FIND(CHAR(1),
SUBSTITUTE(A1,"-",CHAR(1),LEN(A1)-LEN(SUBSTITUTE(A1,"-",""))))
-FIND(CHAR(1),SUBSTITUTE(A1,"\",CHAR(1),LEN(A1)-
LEN(SUBSTITUTE(A1,"\",""))))))
--ron

By the way, here are some UDF's that will do the same thing. They can be
entered in a regular module and then used as a function.

To enter into a regular module, <alt-F11 opens the VBEditor. Ensure your
project is highlighted in the project explorer window, then Insert/Module and
paste one of the codes below into the window that opens:

=============================================
Option Explicit
Function fn(str As String) As String
Dim s1() As String
Dim s2() As String
s1 = Split(str, "\")
s2 = Split(s1(UBound(s1)), "-")
fn = Trim(s2(LBound(s2)))
End Function
==========================================

The above as a "one-liner" in deference to Rick:

============================================
Function fn(str As String) As String
fn = Trim(Split(Split(str, "\")(UBound(Split _
(str, "\"))), "-")(LBound(Split(Split _
(str, "\")(UBound(Split(str, "\"))), "-"))))
End Function
==========================================

and using Regular Expressions, which, although a bit longer, took a fraction of
the time to develop and test of any of the other solutions.

================================
Option Explicit
Function fn(str As String) As String
Dim re As Object, mc As Object
Set re = CreateObject("vbscript.regexp")
re.Pattern = "\\([^\-\\]*\S)\s?-[^\\]*$"
If re.test(str) = True Then
Set mc = re.Execute(str)
fn = mc(0).submatches(0)
End If
End Function
=================================
--ron

On Fri, 15 Feb 2008 22:00:45 -0500, Ron Rosenfeld
wrote:

and using Regular Expressions, which, although a bit longer, took a fraction of
the time to develop and test of any of the other solutions.

================================
Option Explicit
Function fn(str As String) As String
Dim re As Object, mc As Object
Set re = CreateObject("vbscript.regexp")
re.Pattern = "\\([^\-\\]*\S)\s?-[^\\]*$"
If re.test(str) = True Then
Set mc = re.Execute(str)
fn = mc(0).submatches(0)
End If
End Function
=================================

OF course, the regular expression variation as posted above is wrong <g.
Should read:

====================================
Option Explicit
Function fn(str As String) As String
Dim re As Object, mc As Object
Set re = CreateObject("vbscript.regexp")
re.Pattern = "\\([^\\]*\S)\s?-[^\\]*$"
If re.test(str) = True Then
Set mc = re.Execute(str)
fn = mc(0).submatches(0)
End If
End Function
=====================================
--ron

The above as a "one-liner" in deference to Rick:

LOL

============================================
Function fn(str As String) As String
fn = Trim(Split(Split(str, "\")(UBound(Split _
(str, "\"))), "-")(LBound(Split(Split _
(str, "\")(UBound(Split(str, "\"))), "-"))))
End Function
==========================================

The LBound for a Split is always 0 no matter what the Option Base is set to.
Using this fact, your one-liner can be simplified considerably...

Function fn(str As String) As String
fn = Trim(Split(Split(str, "\")(UBound(Split(str, "\"))), "-")(0))
End Function

Rick

On Fri, 15 Feb 2008 23:00:40 -0500, "Rick Rothstein \(MVP - VB\)"
wrote:

The above as a "one-liner" in deference to Rick:

LOL

============================================
Function fn(str As String) As String
fn = Trim(Split(Split(str, "\")(UBound(Split _
(str, "\"))), "-")(LBound(Split(Split _
(str, "\")(UBound(Split(str, "\"))), "-"))))
End Function
==========================================

The LBound for a Split is always 0 no matter what the Option Base is set to.
Using this fact, your one-liner can be simplified considerably...

Function fn(str As String) As String
fn = Trim(Split(Split(str, "\")(UBound(Split(str, "\"))), "-")(0))
End Function



Nice

--ron

On Fri, 15 Feb 2008 23:00:40 -0500, "Rick Rothstein \(MVP - VB\)"
wrote:

The above as a "one-liner" in deference to Rick:

LOL

============================================
Function fn(str As String) As String
fn = Trim(Split(Split(str, "\")(UBound(Split _
(str, "\"))), "-")(LBound(Split(Split _
(str, "\")(UBound(Split(str, "\"))), "-"))))
End Function
==========================================

The LBound for a Split is always 0 no matter what the Option Base is set to.
Using this fact, your one-liner can be simplified considerably...

Function fn(str As String) As String
fn = Trim(Split(Split(str, "\")(UBound(Split(str, "\"))), "-")(0))
End Function


And you've told me that before <grrr <slap upside my head
--ron

On Fri, 15 Feb 2008 23:00:40 -0500, "Rick Rothstein \(MVP - VB\)"
wrote:

The above as a "one-liner" in deference to Rick:

LOL

============================================
Function fn(str As String) As String
fn = Trim(Split(Split(str, "\")(UBound(Split _
(str, "\"))), "-")(LBound(Split(Split _
(str, "\")(UBound(Split(str, "\"))), "-"))))
End Function
==========================================

The LBound for a Split is always 0 no matter what the Option Base is set to.
Using this fact, your one-liner can be simplified considerably...

Function fn(str As String) As String
fn = Trim(Split(Split(str, "\")(UBound(Split(str, "\"))), "-")(0))
End Function

Rick

Actually, neither your one liner nor my longer variants will work if filename
includes a "-". For that, we need something like:

==============================
Option Explicit
Function fn(str As String) As String
Dim s1() As String
s1 = Split(str, "\")
s1 = Split(s1(UBound(s1)), "-")
ReDim Preserve s1(UBound(s1) - 1)
fn = Trim(Join(s1, "-"))
End Function
===========================
--ron

"Ron Rosenfeld" wrote in message
...
On Fri, 15 Feb 2008 23:00:40 -0500, "Rick Rothstein \(MVP - VB\)"
wrote:

The above as a "one-liner" in deference to Rick:

LOL

============================================
Function fn(str As String) As String
fn = Trim(Split(Split(str, "\")(UBound(Split _
(str, "\"))), "-")(LBound(Split(Split _
(str, "\")(UBound(Split(str, "\"))), "-"))))
End Function
==========================================

The LBound for a Split is always 0 no matter what the Option Base is set
to.
Using this fact, your one-liner can be simplified considerably...

Function fn(str As String) As String
fn = Trim(Split(Split(str, "\")(UBound(Split(str, "\"))), "-")(0))
End Function

Rick

Actually, neither your one liner nor my longer variants will work if
filename
includes a "-". For that, we need something like:

==============================
Option Explicit
Function fn(str As String) As String
Dim s1() As String
s1 = Split(str, "\")
s1 = Split(s1(UBound(s1)), "-")
ReDim Preserve s1(UBound(s1) - 1)
fn = Trim(Join(s1, "-"))
End Function
===========================

I thought I had corrected it as per your observation after the OP posted his
sample text line ... use " - ", not "-", in the one-liner...

Function fn(str As String) As String
fn = Trim(Split(Split(str, "\")(UBound(Split(str, "\"))), " - ")(0))
End Function

Of course, this supposes the filename itself does not contain dash
surrounded by spaces.

Rick

Actually, neither your one liner nor my longer variants will work if
filename
includes a "-". For that, we need something like:

==============================
Option Explicit
Function fn(str As String) As String
Dim s1() As String
s1 = Split(str, "\")
s1 = Split(s1(UBound(s1)), "-")
ReDim Preserve s1(UBound(s1) - 1)
fn = Trim(Join(s1, "-"))
End Function
===========================

And, of course, the above would not work if the description itself contained
a dash.

Rick

Actually, neither your one liner nor my longer variants will work if
filename
includes a "-". For that, we need something like:

==============================
Option Explicit
Function fn(str As String) As String
Dim s1() As String
s1 = Split(str, "\")
s1 = Split(s1(UBound(s1)), "-")
ReDim Preserve s1(UBound(s1) - 1)
fn = Trim(Join(s1, "-"))
End Function
===========================

And, of course, the above would not work if the description itself
contained a dash.

In thinking a little more about this question, it would appear, given the
structure the OP has adopted, that **at least one** of the following must be
true or the OP cannot have a fool-proof parser... the filename can never
have a dash, or it can never have a space/dash/space combination in it, or
it can never have just a plain space in it (which would further require a
space always be present after the filename), or the description cannot have
a backslash in it... one of these must be true in order to create a parser
(one-liner or not) that would always work.

Rick

On Sat, 16 Feb 2008 11:09:37 -0500, "Rick Rothstein \(MVP - VB\)"
wrote:

Actually, neither your one liner nor my longer variants will work if
filename
includes a "-". For that, we need something like:

==============================
Option Explicit
Function fn(str As String) As String
Dim s1() As String
s1 = Split(str, "\")
s1 = Split(s1(UBound(s1)), "-")
ReDim Preserve s1(UBound(s1) - 1)
fn = Trim(Join(s1, "-"))
End Function
===========================

And, of course, the above would not work if the description itself
contained a dash.

In thinking a little more about this question, it would appear, given the
structure the OP has adopted, that **at least one** of the following must be
true or the OP cannot have a fool-proof parser... the filename can never
have a dash, or it can never have a space/dash/space combination in it, or
it can never have just a plain space in it (which would further require a
space always be present after the filename), or the description cannot have
a backslash in it... one of these must be true in order to create a parser
(one-liner or not) that would always work.

Rick

That sounds correct.

And, at least for the parser's I've offered, the filename must be preceded by a
"\"
--ron

On Sat, 16 Feb 2008 11:09:37 -0500, "Rick Rothstein \(MVP - VB\)"
wrote:

Actually, neither your one liner nor my longer variants will work if
filename
includes a "-". For that, we need something like:

==============================
Option Explicit
Function fn(str As String) As String
Dim s1() As String
s1 = Split(str, "\")
s1 = Split(s1(UBound(s1)), "-")
ReDim Preserve s1(UBound(s1) - 1)
fn = Trim(Join(s1, "-"))
End Function
===========================

And, of course, the above would not work if the description itself
contained a dash.

In thinking a little more about this question, it would appear, given the
structure the OP has adopted, that **at least one** of the following must be
true or the OP cannot have a fool-proof parser... the filename can never
have a dash, or it can never have a space/dash/space combination in it, or
it can never have just a plain space in it (which would further require a
space always be present after the filename), or the description cannot have
a backslash in it... one of these must be true in order to create a parser
(one-liner or not) that would always work.

Rick

Expanding -- it would certainly be possible for a parser in which the "\"
preceding the filename was optional
--ron

In thinking a little more about this question, it would appear, given the
structure the OP has adopted, that **at least one** of the following must
be
true or the OP cannot have a fool-proof parser... the filename can never
have a dash, or it can never have a space/dash/space combination in it, or
it can never have just a plain space in it (which would further require a
space always be present after the filename), or the description cannot
have
a backslash in it... one of these must be true in order to create a parser
(one-liner or not) that would always work.

Expanding -- it would certainly be possible for a parser in which the "\"
preceding the filename was optional

Which is, of course, a possibility; though, in today's type file
referencing, somewhat rare. If the default path is at the directory where
the file is located, then you can legally specify the file using something
like this... c:filename.ext and the operating system will look in the
current directory. On my system, there is a directory called TEMP at the
root level of my C: drive. In that directory is a file named Test.txt. The
following code (showing a 'backslashless' path reference) prints the first
line of the file into the Immediate window...

Sub test()
ChDir "c:\temp"
Open "c:test.txt" For Input As #1
Line Input #1, LineOfText
Close #1
Debug.Print LineOfText
End Sub

Rick