On Sat, 18 Jul 2009 10:54:43 -0400, "Rick Rothstein"
wrote:

Function First7DigitNumber(S As String) As String
Dim X As Long
For X = 1 To Len(S)
If Mid(" " & S & " ", X, 9) Like "[!0-9]#######[!0-9]" Then
First7DigitNumber = Mid(S, X, 7)
End If
Next
End Function

I note that given the following modification of the OP's test string:

"User requested authority of emergency ID for reason NHUSER1234567 Restore
of object to library LEVEL2 under remedy 61074317. 06/04/09 17:46 QPGMR"

your routine returns 1234567 whereas my UDF returns "MISSING" since there are
no seven digit words.

(Lori's formulas return 1074317)

--ron

I am still not sure if the OP wanted the 7-digit number to stand alone (as a
"word") or not, so I just went for the first isolated 7 digits in a row...

Function First7DigitNumber(S As String) As String
Dim X As Long
For X = 1 To Len(S)
If Mid(" " & S & " ", X, 9) Like "[!0-9]#######[!0-9]" Then
First7DigitNumber = Mid(S, X, 7)
Exit Function
End If
Next
First7DigitNumber = "*MISSING*"
End Function

--
Rick (MVP - Excel)


"Ron Rosenfeld" wrote in message
...
On Sat, 18 Jul 2009 10:54:43 -0400, "Rick Rothstein"
wrote:

Function First7DigitNumber(S As String) As String
Dim X As Long
For X = 1 To Len(S)
If Mid(" " & S & " ", X, 9) Like "[!0-9]#######[!0-9]" Then
First7DigitNumber = Mid(S, X, 7)
End If
Next
End Function

I note that given the following modification of the OP's test string:

"User requested authority of emergency ID for reason NHUSER1234567
Restore
of object to library LEVEL2 under remedy 61074317. 06/04/09 17:46 QPGMR"

your routine returns 1234567 whereas my UDF returns "MISSING" since there
are
no seven digit words.

(Lori's formulas return 1074317)

--ron

My last message had the wrong opening sentence (it was meant for another
message I was working on). This is what I meant to post...

The function I posted does not work as I had indicated... it finds the last
7-digit number, not the first. Here is the corrected code (plus I added the
missing *MISSING* indicator)...

Function First7DigitNumber(S As String) As String
Dim X As Long
For X = 1 To Len(S)
If Mid(" " & S & " ", X, 9) Like "[!0-9]#######[!0-9]" Then
First7DigitNumber = Mid(S, X, 7)
Exit Function
End If
Next
First7DigitNumber = "*MISSING*"
End Function

--
Rick (MVP - Excel)


"Rick Rothstein" wrote in message
...
I am still not sure if the OP wanted the 7-digit number to stand alone (as
a "word") or not, so I just went for the first isolated 7 digits in a
row...

Function First7DigitNumber(S As String) As String
Dim X As Long
For X = 1 To Len(S)
If Mid(" " & S & " ", X, 9) Like "[!0-9]#######[!0-9]" Then
First7DigitNumber = Mid(S, X, 7)
Exit Function
End If
Next
First7DigitNumber = "*MISSING*"
End Function

--
Rick (MVP - Excel)


"Ron Rosenfeld" wrote in message
...
On Sat, 18 Jul 2009 10:54:43 -0400, "Rick Rothstein"
wrote:

Function First7DigitNumber(S As String) As String
Dim X As Long
For X = 1 To Len(S)
If Mid(" " & S & " ", X, 9) Like "[!0-9]#######[!0-9]" Then
First7DigitNumber = Mid(S, X, 7)
End If
Next
End Function

I note that given the following modification of the OP's test string:

"User requested authority of emergency ID for reason NHUSER1234567
Restore
of object to library LEVEL2 under remedy 61074317. 06/04/09 17:46 QPGMR"

your routine returns 1234567 whereas my UDF returns "MISSING" since
there are
no seven digit words.

(Lori's formulas return 1074317)

--ron

I am still not sure if the OP wanted the 7-digit number to stand alone (as a
"word") or not, so I just went for the first isolated 7 digits in a row
whether imbedded in other text or not. The reason I thought that is because
of the "dot" that followed the 7-digit number in the OP's posted example
text. In thinking about it, I'm guessing you took that to be a period at the
end of a sentence.

With that said, I made a mistake in my original function and left out the
*MISSING* indicator. I just posted a corrected function against my original
message for the function.

--
Rick (MVP - Excel)


--
Rick (MVP - Excel)


"Ron Rosenfeld" wrote in message
...
On Sat, 18 Jul 2009 10:54:43 -0400, "Rick Rothstein"
wrote:

Function First7DigitNumber(S As String) As String
Dim X As Long
For X = 1 To Len(S)
If Mid(" " & S & " ", X, 9) Like "[!0-9]#######[!0-9]" Then
First7DigitNumber = Mid(S, X, 7)
End If
Next
End Function

I note that given the following modification of the OP's test string:

"User requested authority of emergency ID for reason NHUSER1234567
Restore
of object to library LEVEL2 under remedy 61074317. 06/04/09 17:46 QPGMR"

your routine returns 1234567 whereas my UDF returns "MISSING" since there
are
no seven digit words.

(Lori's formulas return 1074317)

--ron

I found this definition of a word boundary in Regular Expressions...

"A word boundary represents the spot where a letter or
number meets a space, apostrophe, a period, or anything
else that isn't a letter or number"

Given that, this modification of my function should do what your RegExp
solution does...

Function First7DigitNumber(S As String) As String
Dim X As Long
For X = 1 To Len(S)
If Mid(" " & S & " ", X, 9) Like "[!a-zA-Z0-9]#######[!a-zA-Z0-9]" Then
First7DigitNumber = Mid(S, X, 7)
Exit Function
End If
Next
First7DigitNumber = "*MISSING*"
End Function

--
Rick (MVP - Excel)


"Ron Rosenfeld" wrote in message
...
On Sat, 18 Jul 2009 10:54:43 -0400, "Rick Rothstein"
wrote:

Function First7DigitNumber(S As String) As String
Dim X As Long
For X = 1 To Len(S)
If Mid(" " & S & " ", X, 9) Like "[!0-9]#######[!0-9]" Then
First7DigitNumber = Mid(S, X, 7)
End If
Next
End Function

I note that given the following modification of the OP's test string:

"User requested authority of emergency ID for reason NHUSER1234567
Restore
of object to library LEVEL2 under remedy 61074317. 06/04/09 17:46 QPGMR"

your routine returns 1234567 whereas my UDF returns "MISSING" since there
are
no seven digit words.

(Lori's formulas return 1074317)

--ron

On Sat, 18 Jul 2009 12:33:47 -0400, "Rick Rothstein"
wrote:

The reason I thought that is because
of the "dot" that followed the 7-digit number in the OP's posted example
text. In thinking about it, I'm guessing you took that to be a period at the
end of a sentence.

Yes, I did.


I found this definition of a word boundary in Regular Expressions...

"A word boundary represents the spot where a letter or
number meets a space, apostrophe, a period, or anything
else that isn't a letter or number"

Not quite what I understand it to be (but close)./

The definitions I've seen indicate that a word boundary "Matches at the
position between a word character (anything matched by \w) and a non-word
character (anything matched by [^\w] or \W) as well as at the start and/or end
of the string if the first and/or last characters in the string are word
characters."

And, at least in VBScript, a word character is a digit, letter or underscore
e.g: [A-Za-z0-9_]



Given that, this modification of my function should do what your RegExp
solution does...

Function First7DigitNumber(S As String) As String
Dim X As Long
For X = 1 To Len(S)
If Mid(" " & S & " ", X, 9) Like "[!a-zA-Z0-9]#######[!a-zA-Z0-9]" Then
First7DigitNumber = Mid(S, X, 7)
Exit Function
End If
Next
First7DigitNumber = "*MISSING*"
End Function

It comes pretty close. Just change this line to include the underscores:

If Mid(" " & S & " ", X, 9) Like "[!a-zA-Z0-9_]#######[!a-zA-Z0-9_]" Then

--ron

I had the damnedest time trying to find what RegExp considered a word
boundary (basically, each sight just kept say use \w without listing what it
consider the boundary). When I finally found the one I cited, I figured it
was a universal definition. Now I'm guessing there might be version
differences between the various RegExp engines. Yes, I fix to account for
the underbar is as you have shown it. For the archives, here is the UDF with
the change you indicated...

Function First7DigitNumber(S As String) As String
Dim X As Long
For X = 1 To Len(S)
If Mid(" " & S & " ", X, 9) Like "[!a-zA-Z0-9_]#######[!a-zA-Z0-9_]"
Then
First7DigitNumber = Mid(S, X, 7)
Exit Function
End If
Next
First7DigitNumber = "*MISSING*"
End Function

--
Rick (MVP - Excel)


"Ron Rosenfeld" wrote in message
...
On Sat, 18 Jul 2009 12:33:47 -0400, "Rick Rothstein"
wrote:

The reason I thought that is because
of the "dot" that followed the 7-digit number in the OP's posted example
text. In thinking about it, I'm guessing you took that to be a period at
the
end of a sentence.

Yes, I did.


I found this definition of a word boundary in Regular Expressions...

"A word boundary represents the spot where a letter or
number meets a space, apostrophe, a period, or anything
else that isn't a letter or number"

Not quite what I understand it to be (but close)./

The definitions I've seen indicate that a word boundary "Matches at the
position between a word character (anything matched by \w) and a non-word
character (anything matched by [^\w] or \W) as well as at the start and/or
end
of the string if the first and/or last characters in the string are word
characters."

And, at least in VBScript, a word character is a digit, letter or
underscore
e.g: [A-Za-z0-9_]



Given that, this modification of my function should do what your RegExp
solution does...

Function First7DigitNumber(S As String) As String
Dim X As Long
For X = 1 To Len(S)
If Mid(" " & S & " ", X, 9) Like "[!a-zA-Z0-9]#######[!a-zA-Z0-9]"
Then
First7DigitNumber = Mid(S, X, 7)
Exit Function
End If
Next
First7DigitNumber = "*MISSING*"
End Function

It comes pretty close. Just change this line to include the underscores:

If Mid(" " & S & " ", X, 9) Like "[!a-zA-Z0-9_]#######[!a-zA-Z0-9_]" Then

--ron

On Sat, 18 Jul 2009 19:26:09 -0400, "Rick Rothstein"
wrote:

I had the damnedest time trying to find what RegExp considered a word
boundary (basically, each sight just kept say use \w without listing what it
consider the boundary). When I finally found the one I cited, I figured it
was a universal definition. Now I'm guessing there might be version
differences between the various RegExp engines. Yes, I fix to account for
the underbar is as you have shown it. For the archives, here is the UDF with
the change you indicated...

Function First7DigitNumber(S As String) As String
Dim X As Long
For X = 1 To Len(S)
If Mid(" " & S & " ", X, 9) Like "[!a-zA-Z0-9_]#######[!a-zA-Z0-9_]"
Then
First7DigitNumber = Mid(S, X, 7)
Exit Function
End If
Next
First7DigitNumber = "*MISSING*"
End Function

And, to keep them together, here is the Regex version:

===================
Option Explicit
Function Seven(s As String) As String
Dim re As Object, mc As Object
Set re = CreateObject("vbscript.regexp")
re.Pattern = "\b\d{7}\b"
If re.test(s) = True Then
Set mc = re.Execute(s)
Seven = mc(0)
Else
Seven = "*MISSING*"
End If
End Function
====================

Here are some definitions with regard to word boundaries and some of the
differences:

There are three different positions that qualify as word boundaries:

1. Before the first character in the string, if the first character is
a word character.
2. After the last character in the string, if the last character is a
word character.
3. Between two characters in the string, where one is a word character
and the other is not a word character.

In all flavors, the characters [a-zA-Z0-9_] are word characters. And that is
the case for VBScript. However, some of the other flavors will recognize
characters from other languages, and/or unicode characters, as word characters.
And I believe there is one flavor (?Python) where you can even set flags to
change the definition of a word character.
--ron