Select specific text in cell

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

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

And I meant to my last post that doing this, or any other customization of
parsing, would be fruitless without the OP coming back and filling us in on
what (if any) restrictions exist on his filenames, the descriptions or the
delimiter separating them.

By the way, IF we had to cater to a 'backslashless' path, and IF the
delimiter between the filename and description is in fact a
space/dash/space, the one-liner would become slightly uglier ...

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

Rick

On Sat, 16 Feb 2008 14:52:16 -0500, "Rick Rothstein \(MVP - VB\)"
wrote:

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

And I meant to my last post that doing this, or any other customization of
parsing, would be fruitless without the OP coming back and filling us in on
what (if any) restrictions exist on his filenames, the descriptions or the
delimiter separating them.

By the way, IF we had to cater to a 'backslashless' path, and IF the
delimiter between the filename and description is in fact a
space/dash/space, the one-liner would become slightly uglier ...

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

Rick

Well, if the last "-" is the separator between filename and descriptor, and the
"\" is optional, then the regex is simplified:

re.Pattern = "([^\\]*\S)\s?-[^\\]*$"

As a matter of fact, I think the only circumstance that his regex would fail
would be if there were a "-" within the descriptor.
--ron

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

And I meant to my last post that doing this, or any other customization of
parsing, would be fruitless without the OP coming back and filling us in
on
what (if any) restrictions exist on his filenames, the descriptions or the
delimiter separating them.

By the way, IF we had to cater to a 'backslashless' path, and IF the
delimiter between the filename and description is in fact a
space/dash/space, the one-liner would become slightly uglier ...

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

Rick

Well, if the last "-" is the separator between filename and descriptor,
and the
"\" is optional, then the regex is simplified:

re.Pattern = "([^\\]*\S)\s?-[^\\]*$"

As a matter of fact, I think the only circumstance that his regex would
fail
would be if there were a "-" within the descriptor.

Sorry, but I am not up to speed on my RegEx yet... would that retain the c:
from the front of the path or not?

Rick

On Sat, 16 Feb 2008 14:52:16 -0500, "Rick Rothstein \(MVP - VB\)"
wrote:

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

And I meant to my last post that doing this, or any other customization of
parsing, would be fruitless without the OP coming back and filling us in on
what (if any) restrictions exist on his filenames, the descriptions or the
delimiter separating them.

By the way, IF we had to cater to a 'backslashless' path, and IF the
delimiter between the filename and description is in fact a
space/dash/space, the one-liner would become slightly uglier ...

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

Rick

This, too, will work so long as there is no hyphen within description, but I
couldn't make a one-liner out of it.

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

On Sat, 16 Feb 2008 15:53:05 -0500, "Rick Rothstein \(MVP - VB\)"
wrote:

Sorry, but I am not up to speed on my RegEx yet... would that retain the c:
from the front of the path or not?

If there were no "\" after the C:, then it would be retained. But if the C:
represents a drive, is C:filename legal? OR must it be C:\filename

For the latter, the C:\ would NOT be returned.

--ron

This, too, will work so long as there is no hyphen within description, but
I
couldn't make a one-liner out of it.

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

I think this will work under those conditions...

Function fn(str As String) As String
fn = Trim$(Split(Left(str, InStrRev(str, "-") - 1), _
"\")(UBound(Split(str, "\"))))
End Function

Note that the Trim function call is needed if we are not sure whether the
filename/description delimiter is always a space/dash/space. If that is
always the delimiter, then the function can be simplified to this...

Function fn(str As String) As String
fn = Split(Left(str, InStrRev(str, "-") - 2), _
"\")(UBound(Split(str, "\")))
End Function

By the way, I estimate that statement line is 3 characters too long to fit
on one, non-continued line before newsreader line wrapping would mangle it;
hence, the line continuation. For example, if we reduce the argument name to
just S, then the function is this neater looking one...

Function fn(S As String) As String
fn = Split(Left(S, InStrRev(S, "-") - 2), "\")(UBound(Split(S, "\")))
End Function

Rick

Sorry, but I am not up to speed on my RegEx yet... would that retain the
c:
from the front of the path or not?

If there were no "\" after the C:, then it would be retained. But if the
C:
represents a drive, is C:filename legal? OR must it be C:\filename

Yes, c:filename is legal. In the flurry of postings we have done, you missed
my 2:40PM (EST) message. I am repeating it here so you don't have to look
for it... see my test() subroutine and note in particular the path/filename
used in the Open statement.

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

On Sat, 16 Feb 2008 16:31:39 -0500, "Rick Rothstein \(MVP - VB\)"
wrote:

Sorry, but I am not up to speed on my RegEx yet... would that retain the
c:
from the front of the path or not?

If there were no "\" after the C:, then it would be retained. But if the
C:
represents a drive, is C:filename legal? OR must it be C:\filename

Yes, c:filename is legal. In the flurry of postings we have done, you missed
my 2:40PM (EST) message. I am repeating it here so you don't have to look
for it... see my test() subroutine and note in particular the path/filename
used in the Open statement.

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

Well, as written, the regex would retain the C:.

If you wanted to return filename without the C:, it would be a simple
alteration in the regex:

re.Pattern = "([^\\:]*\S)\s?-[^\\]*$"

This is a bit more robust, though:

re.Pattern = "([^\\:]*\S)\s*-[^\\]*$"


--ron