Help with Regexp, please

Hi,

The regular expression (\d{15,16}) matches a substring in a cell. I
want to extract the remaining part of the cell ie. from the character
after the matched substring till the end of the string in the cell
using a regular expression.

Is it possible to do this?

Thanks in advance for the help.

Regards,
Raj

0
Raj
5/1/2010 12:39:32 PM
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Raj,

with
abcd15,16xyz
in a cell the code below abstracts
abcxyz.
Maybe you can use this as a  basis to develop your code

Function GetSubstring(S As String) As String
 Dim re As Object
 Set re = CreateObject("vbscript.regexp")
    re.Pattern = "[d{15,16}]"
    re.Global = True
    GetSubstring = re.Replace(S, "")
End Function


"Raj" <rspai9@gmail.com> wrote in message
news:e48052c8-66d1-4508-ab52-a307be52579a@a39g2000prb.googlegroups.com...
> Hi,
>
> The regular expression (\d{15,16}) matches a substring in a cell. I
> want to extract the remaining part of the cell ie. from the character
> after the matched substring till the end of the string in the cell
> using a regular expression.
>
> Is it possible to do this?
>
> Thanks in advance for the help.
>
> Regards,
> Raj
>


0
Project
5/1/2010 1:13:06 PM
Raj,

Posted this earlier but it has disappeared into cyberspace:

with
abcd15,16xyz
in a cell the code below abstracts
abcxyz.
Maybe you can use this as a  basis to develop your code

Function GetSubstring(S As String) As String
 Dim re As Object
 Set re = CreateObject("vbscript.regexp")
    re.Pattern = "[d{15,16}]"
    re.Global = True
    GetSubstring = re.Replace(S, "")
End Function


HTH


"Raj" <rspai9@gmail.com> wrote in message
news:e48052c8-66d1-4508-ab52-a307be52579a@a39g2000prb.googlegroups.com...
> Hi,
>
> The regular expression (\d{15,16}) matches a substring in a cell. I
> want to extract the remaining part of the cell ie. from the character
> after the matched substring till the end of the string in the cell
> using a regular expression.
>
> Is it possible to do this?
>
> Thanks in advance for the help.
>
> Regards,
> Raj
>


0
Project
5/1/2010 5:17:17 PM
On Sat, 1 May 2010 05:39:32 -0700 (PDT), Raj <rspai9@gmail.com> wrote:

>Hi,
>
>The regular expression (\d{15,16}) matches a substring in a cell. I
>want to extract the remaining part of the cell ie. from the character
>after the matched substring till the end of the string in the cell
>using a regular expression.
>
>Is it possible to do this?
>
>Thanks in advance for the help.
>
>Regards,
>Raj

(\d{15,16})([\s\S]*)

will capture everything in the cell and after your "match" into Group 2




--ron
0
Ron
5/1/2010 6:47:46 PM
On Sat, 01 May 2010 14:47:46 -0400, Ron Rosenfeld
<ronrosenfeld@nospam.org> wrote:

>On Sat, 1 May 2010 05:39:32 -0700 (PDT), Raj <rspai9@gmail.com> wrote:
>
>>Hi,
>>
>>The regular expression (\d{15,16}) matches a substring in a cell. I
>>want to extract the remaining part of the cell ie. from the character
>>after the matched substring till the end of the string in the cell
>>using a regular expression.
>>
>>Is it possible to do this?
>>
>>Thanks in advance for the help.
>>
>>Regards,
>>Raj
>
>(\d{15,16})([\s\S]*)
>
>will capture everything in the cell and after your "match" into Group 2
>
>
>
>
>--ron

Can't [\s\S] be replaced by . like this
(\d{15,16})(.*)

Lars-�ke

0
Lars
5/1/2010 7:08:01 PM
To clarify, the regex should return " Raffles Traders" from the string
below:

NRK2D	986123456789312 Raffles Traders

The regex (\d{15,16})([\s\S]*) is returning "986123456789312 Raffles
Traders"

Regards,
Raj



On May 2, 12:08=A0am, Lars-=C5ke Aspelin <lar...@REMOOVEtelia.com> wrote:
> On Sat, 01 May 2010 14:47:46 -0400, Ron Rosenfeld
>
>
>
> <ronrosenf...@nospam.org> wrote:
> >On Sat, 1 May 2010 05:39:32 -0700 (PDT), Raj <rsp...@gmail.com> wrote:
>
> >>Hi,
>
> >>The regular expression (\d{15,16}) matches a substring in a cell. I
> >>want to extract the remaining part of the cell ie. from the character
> >>after the matched substring till the end of the string in the cell
> >>using a regular expression.
>
> >>Is it possible to do this?
>
> >>Thanks in advance for the help.
>
> >>Regards,
> >>Raj
>
> >(\d{15,16})([\s\S]*)
>
> >will capture everything in the cell and after your "match" into Group 2
>
> >--ron
>
> Can't [\s\S] be replaced by . like this
> (\d{15,16})(.*)
>
> Lars-=C5ke

0
Raj
5/1/2010 8:41:03 PM
On Sat, 01 May 2010 21:08:01 +0200, Lars-�ke Aspelin <larske@REMOOVEtelia.com>
wrote:

>Can't [\s\S] be replaced by . like this
>(\d{15,16})(.*)
>
>Lars-�ke

Your suggestion will work IF and ONLY IF there are no line feeds or carriage
returns in the cell.  In some flavors, there is an option to have Dot match
newline, but such does not exist in VBA (or Javascript).

If the OP, rather than wanting to extract everything to " ... the end of the
string in the cell"  only wanted to extract everything to the end of the line,
and ignore anything in the cell after a newline character, then (.*) would be
appropriate.
--ron
0
Ron
5/1/2010 8:43:48 PM
I have made another post before Ron's last post. I have explained the
requirement with an example.

Regards,
Raj



On May 2, 1:43=A0am, Ron Rosenfeld <ronrosenf...@nospam.org> wrote:
> On Sat, 01 May 2010 21:08:01 +0200, Lars-=C5ke Aspelin <lar...@REMOOVEtel=
ia.com>
> wrote:
>
> >Can't [\s\S] be replaced by . like this
> >(\d{15,16})(.*)
>
> >Lars-=C5ke
>
> Your suggestion will work IF and ONLY IF there are no line feeds or carri=
age
> returns in the cell. =A0In some flavors, there is an option to have Dot m=
atch
> newline, but such does not exist in VBA (or Javascript).
>
> If the OP, rather than wanting to extract everything to " ... the end of =
the
> string in the cell" =A0only wanted to extract everything to the end of th=
e line,
> and ignore anything in the cell after a newline character, then (.*) woul=
d be
> appropriate.
> --ron

0
Raj
5/1/2010 9:08:24 PM
Just in case you are interested, here is some non-RegEx code that finds the 
same part of your text (the result is returned in the variable named 
TailEnd)...

Text = "NRK2D 986123456789312 Raffles Traders"
For X = 1 To Len(Text)
  If Mid(Text, X, 15) Like String(15, "#") Then
    TailEnd = Mid(Text, X + 15)
    Exit For
  End If
Next

-- 
Rick (MVP - Excel)



"Raj" <rspai9@gmail.com> wrote in message 
news:c6942365-a8c5-4208-bb88-bf91e0628b05@p5g2000pri.googlegroups.com...
> To clarify, the regex should return " Raffles Traders" from the string
> below:
>
> NRK2D 986123456789312 Raffles Traders
>
> The regex (\d{15,16})([\s\S]*) is returning "986123456789312 Raffles
> Traders"
>
> Regards,
> Raj
>
>
>
> On May 2, 12:08 am, Lars-�ke Aspelin <lar...@REMOOVEtelia.com> wrote:
>> On Sat, 01 May 2010 14:47:46 -0400, Ron Rosenfeld
>>
>>
>>
>> <ronrosenf...@nospam.org> wrote:
>> >On Sat, 1 May 2010 05:39:32 -0700 (PDT), Raj <rsp...@gmail.com> wrote:
>>
>> >>Hi,
>>
>> >>The regular expression (\d{15,16}) matches a substring in a cell. I
>> >>want to extract the remaining part of the cell ie. from the character
>> >>after the matched substring till the end of the string in the cell
>> >>using a regular expression.
>>
>> >>Is it possible to do this?
>>
>> >>Thanks in advance for the help.
>>
>> >>Regards,
>> >>Raj
>>
>> >(\d{15,16})([\s\S]*)
>>
>> >will capture everything in the cell and after your "match" into Group 2
>>
>> >--ron
>>
>> Can't [\s\S] be replaced by . like this
>> (\d{15,16})(.*)
>>
>> Lars-�ke
> 
0
Rick
5/1/2010 9:32:04 PM
On Sat, 1 May 2010 13:41:03 -0700 (PDT), Raj <rspai9@gmail.com> wrote:

>To clarify, the regex should return " Raffles Traders" from the string
>below:
>
>NRK2D	986123456789312 Raffles Traders
>
>The regex (\d{15,16})([\s\S]*) is returning "986123456789312 Raffles
>Traders"
>
>Regards,
>Raj

Obviously, you are not doing what I suggested which was to return the *SECOND*
matching group.  You are returning the ENTIRE match.  Here's an example as to
returning the *SECOND* match using VBA:

===============================
Option Explicit
Function Part2(s As String) As String
 Dim re As Object, mc As Object
Set re = CreateObject("vbscript.regexp")
    re.Pattern = "(\d{15,16})([\s\S]*)"
    If re.test(s) = True Then
        Set mc = re.Execute(s)
        Part2 = mc(0).submatches(1)
    End If
End Function
==================================
--ron
0
Ron
5/1/2010 11:17:52 PM
On Sat, 1 May 2010 17:32:04 -0400, "Rick Rothstein"
<rick.newsNO.SPAM@NO.SPAMverizon.net> wrote:

>Just in case you are interested, here is some non-RegEx code that finds the 
>same part of your text (the result is returned in the variable named 
>TailEnd)...
>
>Text = "NRK2D 986123456789312 Raffles Traders"
>For X = 1 To Len(Text)
>  If Mid(Text, X, 15) Like String(15, "#") Then
>    TailEnd = Mid(Text, X + 15)
>    Exit For
>  End If
>Next
>
>-- 
>Rick (MVP - Excel)

Rick,

The OP's original regex (\d{15,16}) will capture the first 15 *OR 16* digit
string into capture group 1.  So when the *rest* of the string is returned, it
will omit digit 16 if present.

Your code will return the 16th digit, if present, as a part of "TailEnd"
--ron
0
Ron
5/1/2010 11:52:19 PM
Okay, I wasn't entirely sure what the "15,16" meant in the pattern. If the 
numbers are always followed by a space (is that what the "\s\S" part of your 
expression is for?), my code could be modified to this...

Text = "NRK2D 3298613456378931 Raffles Traders"
For X = 1 To Len(Text)
  If Mid(Text, X, 16) Like String(15, "#") & " " Then
    TailEnd = Mid(Text, X + 15)
    Exit For
  End If
Next

-- 
Rick (MVP - Excel)



"Ron Rosenfeld" <ronrosenfeld@nospam.org> wrote in message 
news:9hfpt5t32sem56v3n5ttmjfgbooo8mvta7@4ax.com...
> On Sat, 1 May 2010 17:32:04 -0400, "Rick Rothstein"
> <rick.newsNO.SPAM@NO.SPAMverizon.net> wrote:
>
>>Just in case you are interested, here is some non-RegEx code that finds 
>>the
>>same part of your text (the result is returned in the variable named
>>TailEnd)...
>>
>>Text = "NRK2D 986123456789312 Raffles Traders"
>>For X = 1 To Len(Text)
>>  If Mid(Text, X, 15) Like String(15, "#") Then
>>    TailEnd = Mid(Text, X + 15)
>>    Exit For
>>  End If
>>Next
>>
>>-- 
>>Rick (MVP - Excel)
>
> Rick,
>
> The OP's original regex (\d{15,16}) will capture the first 15 *OR 16* 
> digit
> string into capture group 1.  So when the *rest* of the string is 
> returned, it
> will omit digit 16 if present.
>
> Your code will return the 16th digit, if present, as a part of "TailEnd"
> --ron 

0
Rick
5/2/2010 12:19:07 AM
On Sat, 1 May 2010 20:19:07 -0400, "Rick Rothstein"
<rick.newsNO.SPAM@NO.SPAMverizon.net> wrote:

>Okay, I wasn't entirely sure what the "15,16" meant in the pattern. If the 
>numbers are always followed by a space (is that what the "\s\S" part of your 
>expression is for?), my code could be modified to this...
>

Whatever it is that follows the string of digits is undefined in the OP's
specifications.  It could be a space; it could be another digit; it could be an
alpha character; it could be nothing.  That is a problem with the
specification.

Also, whatever it is that precedes the string of digits is ALSO unspecified. It
could even be another digit!

If, as in the OP's example, the string of digits is ALWAYS preceded and
followed by a space, then his regex should have been something like:

\s\d{15,16}\s

If he only wanted to capture the standalone string of digits, then

\s(\d{15,16})\s  would capture just the digits into Group 1.

and, expanding on that,

\s\d{15,16}\s+([\s\S]*)  would also
	capture everything after the string of digits into group 2 except for
the leading <space>'s before group 2.

In my suggestion, the [\s\S] will match every character that is either a
<space> or not a <space>.  In other words, it captures everything.

If all I wanted to do was return everything in the string that came after a 15
or 16 digit number, that was bounded by spaces, I would just replace the
beginning of the string with nothing.  It's much simpler, and probably faster.
===================================
Option Explicit
Function Part2(s As String) As String
 Dim re As Object
Set re = CreateObject("vbscript.regexp")
 re.Pattern = "^[\s\S]+\s\d{15,16}\s+"
 Part2 = re.Replace(s, "")
End Function
====================================

If I wanted to include the space prior to " Raffles Traders" as the OP did in
his example, then perhaps I would use this pattern:

re.Pattern = "^[\s\S]+\s\d{15,16}\b"

Now this would return the unaltered string if there was no match, but we could
easily test for that, depending on what the OP wanted to do in that instance.

-----------------
if re.test(s) = true then
	part2 = re.replace(s,"")
else
	part2 = "no pattern match"
end if
--------------------
--ron
0
Ron
5/2/2010 12:58:47 AM
Thanks for the solved problem and the learning about submatches that I
was not aware of.

Regards,
Raj




On May 2, 5:58=A0am, Ron Rosenfeld <ronrosenf...@nospam.org> wrote:
> On Sat, 1 May 2010 20:19:07 -0400, "Rick Rothstein"
>
> <rick.newsNO.S...@NO.SPAMverizon.net> wrote:
> >Okay, I wasn't entirely sure what the "15,16" meant in the pattern. If t=
he
> >numbers are always followed by a space (is that what the "\s\S" part of =
your
> >expression is for?), my code could be modified to this...
>
> Whatever it is that follows the string of digits is undefined in the OP's
> specifications. =A0It could be a space; it could be another digit; it cou=
ld be an
> alpha character; it could be nothing. =A0That is a problem with the
> specification.
>
> Also, whatever it is that precedes the string of digits is ALSO unspecifi=
ed. It
> could even be another digit!
>
> If, as in the OP's example, the string of digits is ALWAYS preceded and
> followed by a space, then his regex should have been something like:
>
> \s\d{15,16}\s
>
> If he only wanted to capture the standalone string of digits, then
>
> \s(\d{15,16})\s =A0would capture just the digits into Group 1.
>
> and, expanding on that,
>
> \s\d{15,16}\s+([\s\S]*) =A0would also
> =A0 =A0 =A0 =A0 capture everything after the string of digits into group =
2 except for
> the leading <space>'s before group 2.
>
> In my suggestion, the [\s\S] will match every character that is either a
> <space> or not a <space>. =A0In other words, it captures everything.
>
> If all I wanted to do was return everything in the string that came after=
 a 15
> or 16 digit number, that was bounded by spaces, I would just replace the
> beginning of the string with nothing. =A0It's much simpler, and probably =
faster.
> =3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D
> Option Explicit
> Function Part2(s As String) As String
> =A0Dim re As Object
> Set re =3D CreateObject("vbscript.regexp")
> =A0re.Pattern =3D "^[\s\S]+\s\d{15,16}\s+"
> =A0Part2 =3D re.Replace(s, "")
> End Function
> =3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D
>
> If I wanted to include the space prior to " Raffles Traders" as the OP di=
d in
> his example, then perhaps I would use this pattern:
>
> re.Pattern =3D "^[\s\S]+\s\d{15,16}\b"
>
> Now this would return the unaltered string if there was no match, but we =
could
> easily test for that, depending on what the OP wanted to do in that insta=
nce.
>
> -----------------
> if re.test(s) =3D true then
> =A0 =A0 =A0 =A0 part2 =3D re.replace(s,"")
> else
> =A0 =A0 =A0 =A0 part2 =3D "no pattern match"
> end if
> --------------------
> --ron

0
Raj
5/2/2010 2:52:40 PM
On Sun, 2 May 2010 07:52:40 -0700 (PDT), Raj <rspai9@gmail.com> wrote:

>Thanks for the solved problem and the learning about submatches that I
>was not aware of.
>
>Regards,
>Raj

Glad to help.  Thanks for the feedback.

In your initial posting, you had your regex within parentheses:

	(\d{15,16})
	^         ^

That captures that result into a capturing group, which, in VBA, is referenced
as a submatch.  That was why I assumed that you were aware of that concept.  I
should have been more explicit.
--ron
0
Ron
5/2/2010 3:27:43 PM
Reply:

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I'm looking at a website on CString Management: http://www.codeproject.com:80/string/cstringmgmt.asp In the section entitled, "CString to char * II: Using GetBuffer," the author stresses calling ReleaseBuffer after calling GetBuffer. Is this always necessary? I often use CString::GetBuffer when using CStrings in MessageBox dialogs like so: MessageBox(m_hWnd, cString.GetBuffer(0), lpTitle, MB_OK); Should I be adding a ReleaseBuffer after a MessageBox call? Should I be passing my string data to the MessageBox in another way? Regards, Joe > Is this always necessary?...

HelpProvider and HTML Help interaction
I have an application with a .chm help-file. But I have some questions about the behaviour of the help-window. I use a modal application window and I can start the help. The help windows appears but it is allways in foreground of my application window. I can set the input focus on my window, the help window becomes inactive but I cannot move my window over the help window. So I have to close the help window or minimize it or move it aside of my application window when I want to go on in my application. The other problem is when I minimize the help window and then open a dialog...

Your Help is Appreciated
Dear all I am using MS Excel 2003 and I am trying to plot a graph/chart/graphical representation of the following: I would like to somehow plot : Time Price Meetings 16/09/2002 23/09/2002 225 27-Sep-02 30/09/2002 07/10/2002 14/10/2002 368 21/10/2002 23-Oct-02 28/10/2002 04/11/2002 354 11/11/2002 18/11/2002 235 25/11/2002 02/12/2002 I dont mind how the data is displayed but I envisaged some kind of bar chart to represent the prices and a line chart / crosses to indicate the dates of the meetings. Can anyone help with this or am I being stupid? Thanks for everything i...

Money-Changing Account number to handle Vanguard changes
Hi: Vanguard has changed the account numbers for its brokerage accounts. This has caused money to download brand new accounts with nothing in them and stop updating my old account numbered accounts. I changed the account numbers in Account Detail page but still not working. Anyone who has any info on what I'm missing please pass along. I cannot lose years and years of past data by just switching to the newly downloaded accounts. Greg PS: Microsoft Money Plus Premium, most recent version On the old accounts, turn off on-line updates. You should then be able to merge the old accoun...

bcc help
can't seem to send mail to bcc? I copied e-mail addresses from an excel spreadsheet and pasted the into the bcc header. I made sure there was coma between each. I entered the form letter and "from" and "to" and "subject" fields an sent the mail. The "to" received the mail but the bcc did NOT receive the mail. What did I do wrong ----------------------------------------------- ~~ Message posted from http://www.ExcelTip.com ~~View and post usenet messages directly from http://www.ExcelForum.com ...

Rules Wizard Help
I have a rule in the Rules Wizard that forwards on an email to distribution list that is received from a certain person. I would like to have a rule that forwards on this email, but I woul like to forward it with some set text as well. For example, the email shows after being forwarded by the rule: > ---------- > From: Person1 > Sent: Friday, February 20, 2004 6:52:37 AM > To: DistributionList > Subject: DISCOVERER SHOULD NOW BE AVAILABLE > Auto forwarded by a Rule > > DISCOVERER SHOULD NOW BE AVAILABLE However, after I would like it to forward some text as ...

More help with formula
I need 2 formulas for the following: Problem 1: Columns "c" thru "g" contain numbers from 1 thru 100. If th numbers is those columns are 50 or less then I the sum to go in on column; then the sum of numbers between 51 to 100 to go into anothe column. Problem 2: Columns "c" thru "g" contain numbers from 1 thru 100. If th individual number 1-100 is used (3 times, 4 times, 5 times, etc.) tota the I need to show the number of times it was used in the columns. For example: If 1 is used 5 times in columns C thru G then the numbe is 5; the same with 2, th...

help help help IF function
i am new to excell and need help with a problem...... the problem is iam looking for a amount < = 3000.00 and iam to display in bold and in light green i have never had to deal with that i got as far as =IF<=3000 then iam lost or it that even right?? I'm far from a whiz at Excel but I'd use conditional formatting. -- JoAnn Paules MVP Microsoft [Publisher] ~~~~~ How to ask a question http://support.microsoft.com/KB/555375 "kimmy" <kimmy@discussions.microsoft.com> wrote in message news:ADC77884-FC87-47A3-90F2-D2AAF957F5D1@microsoft.com... >i a...

IF OR help
I have entered the formula =IF(OR(D5="K",D5<=4,M32<=95),"INELIGIBLE"," ") I am trying to get a result that will display INELIGIBLE when the value in D5 is K,1,2,3,4 OR the value in M32 is less than 96. with this formula I now get INELIGIBLE when D5 is blank. I would like cell to be blank unless one of the specific conditions is met. If either D5 or M32 is blank, I would like this cell to also be blank M32 contains a formula that returns a blank cell unless data is entered elsewhere in the workbook Thanks for all the help -- dbconn...

Help -- Help -- Help
Hi, I am having a lot of trouble with Outlook. I understand the other Microsoft programs, but for some reason I can't get Outlook. Does anyone know of any books that really breakdown the concepts of Outlook? Thank you. Donald Donald8044 wrote: > Hi, > > I am having a lot of trouble with Outlook. > > I understand the other Microsoft programs, but for some reason I > can't get Outlook. > > Does anyone know of any books that really breakdown the concepts of > Outlook? > > Thank you. > > Donald Umm what exactly do you mean - "but for some...

Drop Down Menu Help #2
Leave a blank row? How would I do that? For the drop down, I use th validation function, and it lets you list what you put in to the dro down and you have to separate it by commas. a,b,c,d, ect. How would do what you are suggesting -- mcr ----------------------------------------------------------------------- mcr1's Profile: http://www.excelforum.com/member.php?action=getinfo&userid=1549 View this thread: http://www.excelforum.com/showthread.php?threadid=27475 Then you follow Debra's suggestion. I took for granted that you were using a list contained in a named range, where ...

Help, please: sorting the Legend
I'm lucky if I use 10% of the power of Excel. I would be grateful for advice. Ten salesmen on a team. At the end of each month, their cumulative totals for the year are updated. The spreadsheet gets re-sorted, showing their standing in rank order. Simple enough. At the beginning of the year, I created a line chart to track each person's progress. I initially listed the names in alphabetical order and defined the series accordingly. The problem comes when I do the end of the month re-sort. The names in the Legend are no longer in alphabetical order but neither do they correctly corres...