R
Ron Rosenfeld
Expanding -- it would certainly be possible for a parser in which the "\"
preceding the filename was optional
--ron
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Expanding -- it would certainly be possible for a parser in which the "\"
preceding the filename was optional
--ron
R
Rick Rothstein \(MVP - VB\)
In thinking a little more about this question, it would appear, given the
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
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
R
Rick Rothstein \(MVP - VB\)
Expanding -- it would certainly be possible for a parser in which
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
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
R
Ron Rosenfeld
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
R
Rick Rothstein \(MVP - VB\)
Expanding -- it would certainly be possible for a parser in which
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
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
R
Ron Rosenfeld
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
R
Ron Rosenfeld
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
R
Rick Rothstein \(MVP - VB\)
This, too, will work so long as there is no hyphen within description, but
I think this will work under those conditions...
Function fn(str As String) As String
fn = Trim$(Split(Left(str, InStrRev(str, "-") - 1), _
"\")(UBound(Split(str, "\"))))
End Function
Note that the Trim function call is needed if we are not sure whether the
filename/description delimiter is always a space/dash/space. If that is
always the delimiter, then the function can be simplified to this...
Function fn(str As String) As String
fn = Split(Left(str, InStrRev(str, "-") - 2), _
"\")(UBound(Split(str, "\")))
End Function
By the way, I estimate that statement line is 3 characters too long to fit
on one, non-continued line before newsreader line wrapping would mangle it;
hence, the line continuation. For example, if we reduce the argument name to
just S, then the function is this neater looking one...
Function fn(S As String) As String
fn = Split(Left(S, InStrRev(S, "-") - 2), "\")(UBound(Split(S, "\")))
End Function
Rick
I
couldn't make a one-liner out of it.
==========================
Option Explicit
Function fn(str As String) As String
Dim s1() As String
s1 = Split(Split(str, "\")(UBound(Split(str, "\"))), "-")
ReDim Preserve s1(UBound(s1) - 1)
fn = Join(s1, "-")
End Function
==============================
I think this will work under those conditions...
Function fn(str As String) As String
fn = Trim$(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
R
Rick Rothstein \(MVP - VB\)
Sorry, but I am not up to speed on my RegEx yet... would that retain the
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.
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
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
R
Ron Rosenfeld
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.
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
R
Rick Rothstein \(MVP - VB\)
Sorry, but I am not up to speed on my RegEx yet... would that retain the
I wouldn't worry about it. Even though it is legal to have a path with no
backslashes, almost no one does so anymore as it requires some mechanism to
change the active directory path in order to use it. One would hardly store
paths in that format with the requirement that to use them, a certain path
has to be made active.
Rick
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.
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?-[^\\]*$"
This is a bit more robust, though:
re.Pattern = "([^\\:]*\S)\s*-[^\\]*$"
I wouldn't worry about it. Even though it is legal to have a path with no
backslashes, almost no one does so anymore as it requires some mechanism to
change the active directory path in order to use it. One would hardly store
paths in that format with the requirement that to use them, a certain path
has to be made active.
Rick
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