Help with MID Function?

[KLZA]

Hi. I may not have explained myself well enough in my last post. So
here goes again.
I'm trying to capture / extract specific data within a string of text
in a cell. I need to capture a string of numeric characters plus only
the first alpha character immediately after the numbers. The numbers
of alpha characters varies before the numeric characters. So my cells
could look like this: TTTTT10TTTTTT or TTT1000TTTT or
TTTTTTT1TTTTTTTTTTTT etc.. My output from the cells would be 10T or
1000T or 1T etc... The text string before and after the number string
vary. Can this be done?

 

[Rick Rothstein]

Sure... just use the last formula I posted back in your first thread on this
problem; namely,

=MID(A1,MIN(FIND({0,1,2,3,4,5,6,7,8,9},A1&"0123456789")),LEN(LOOKUP(9E+307,--LEFT(MID(A1,MIN(FIND({0,1,2,3,4,5,6,7,8,9},A1&"0123456789")),999),ROW(1:999))))+1)

Change all the A1 references (4 of them) to whatever cell you have your text
in. This formula can be copied down if necessary.

 

[KLZA]

Can you describe what "doesn't seem to work" means? What do you see the
formula returning? Did you change the cell references (the A1's) to the cell
that has your text?

Try this experiment. Put one of your text strings in A1; then click in A3to
activate it (really, any cell will do); and then copy/paste the formula I
posted (do not try to re-type it) into the Formula Bar. and hit Enter. What
is displayed in A3?

--
Rick (MVP - Excel)





Hi.  First, thanks for the help.  The formula doesn't seem to work.
Can someone test it or explain what I'm possibly doing wrong?- Hide quoted text -

- Show quoted text -

I get an error after 9E+307,

 

[Rick Rothstein]

I'm not sure what to tell you... I tested the formula before I posted it and
I just tested it again and it works on my XL2003 worksheet without any
problems. You did copy/paste the formula into your Formula Bar, didn't you?

--
Rick (MVP - Excel)

Can you describe what "doesn't seem to work" means? What do you see the
formula returning? Did you change the cell references (the A1's) to the
cell
that has your text?

Try this experiment. Put one of your text strings in A1; then click in A3
to
activate it (really, any cell will do); and then copy/paste the formula I
posted (do not try to re-type it) into the Formula Bar. and hit Enter.
What
is displayed in A3?

--
Rick (MVP - Excel)





Hi. First, thanks for the help. The formula doesn't seem to work.
Can someone test it or explain what I'm possibly doing wrong?- Hide quoted
text -

- Show quoted text -

I get an error after 9E+307,

 

[KLZA]

I'm not sure what to tell you... I tested the formula before I posted it and
I just tested it again and it works on my XL2003 worksheet without any
problems. You did copy/paste the formula into your Formula Bar, didn't you?

--
Rick (MVP - Excel)







I get an error after 9E+307,- Hide quoted text -

- Show quoted text -

Hey, I tested on two other machines and I think your formula may be
faulty. I'm using excel 2003 with different builds.. Can someone
else test this?

 

[Ron Rosenfeld]

Hi. I may not have explained myself well enough in my last post. So
here goes again.
I'm trying to capture / extract specific data within a string of text
in a cell. I need to capture a string of numeric characters plus only
the first alpha character immediately after the numbers. The numbers
of alpha characters varies before the numeric characters. So my cells
could look like this: TTTTT10TTTTTT or TTT1000TTTT or
TTTTTTT1TTTTTTTTTTTT etc.. My output from the cells would be 10T or
1000T or 1T etc... The text string before and after the number string
vary. Can this be done?

Repeating my last post:

Easy to do with a UDF:

To enter this <alt-F11> opens the VB Editor. Ensure your project is
highlighted in the Project Explorer window, then Insert/Module and paste the
code below into the window that opens.

To use this, enter a formula of the type:

=extrNumsPlusOne(A1)

The pattern (in the UDF below) will find the first series of digits and a
following single alpha character.

==============================
Function extrNumsPlusOne(str As String) As String
Dim re As Object, mc As Object
Set re = CreateObject("vbscript.regexp")
re.Pattern = "\d+[A-Za-z]"
If re.test(str) = True Then
Set mc = re.Execute(str)
extrNumsPlusOne = mc(0).Value
End If
End Function
============================
--ron
--ron

 

[JeffP->]

Here's a VBA example, not as neat as a worksheet function but here ya go...

Cells A1:I1
TTTT100TT
cell J1 enter =GetNumAndChar(A1:I1)

Dragging the lower right corner of this cell to other locations create the
referrenced incremented ranges.

HTH

<begin copy omit this line>
Option Explicit

Public Function GetNumAndChar(ByVal rRange As Excel.Range) As String

Dim in_value As String
Dim iRow, iCol, iposit, iStop As Integer
in_value = ""
iRow = 1

For iCol = 1 To rRange.Cells.Count
If rRange.Cells(iRow, iCol) = "." Then iCol = iCol + 1
If IsNumeric(rRange.Cells(iRow, iCol)) Then
in_value = in_value & rRange.Cells(iRow, iCol)
If IsNumeric(rRange.Cells(iRow, iCol)) And Not
IsNumeric(rRange.Cells(iRow, iCol + 1)) Then
in_value = in_value & rRange.Cells(iRow, iCol + 1)
Exit For
End If
End If
Next

If IsNull(in_value) Then
GetNumAndChar = " "
Else: GetNumAndChar = in_value
End If

End Function
<end copy omit this line>

 

[Rick Rothstein]

You are right... there is a problem with the formula IF your number starts
with one or more zeroes. I'll look into trying to correct it.

--
Rick (MVP - Excel)

I'm not sure what to tell you... I tested the formula before I posted it
and
I just tested it again and it works on my XL2003 worksheet without any
problems. You did copy/paste the formula into your Formula Bar, didn't
you?

--
Rick (MVP - Excel)







I get an error after 9E+307,- Hide quoted text -

- Show quoted text -

Hey, I tested on two other machines and I think your formula may be
faulty. I'm using excel 2003 with different builds.. Can someone
else test this?

 

[KLZA]

Repeating my last post:

Easy to do with a UDF:

To enter this <alt-F11> opens the VB Editor.  Ensure your project is
highlighted in the Project Explorer window, then Insert/Module and paste the
code below into the window that opens.

To use this, enter a formula of the type:

The pattern (in the UDF below) will find the first series of digits anda
following single alpha character.

==============================
Function extrNumsPlusOne(str As String) As String
Dim re As Object, mc As Object
Set re = CreateObject("vbscript.regexp")
re.Pattern = "\d+[A-Za-z]"
If re.test(str) = True Then
    Set mc = re.Execute(str)
    extrNumsPlusOne = mc(0).Value
End If
End Function
============================
--ron
--ron

Fantastic!  You da man!  This works great!  12800 line items - done!- Hide quoted text -

- Show quoted text -

How do I modify to extract only the number?

 

[Ron Rosenfeld]

Glad to help. Thanks for the feedback.
--ron

 

[Ron Rosenfeld]

=extrNumsPlusOne(A1)

The pattern (in the UDF below) will find the first series of digits and a
following single alpha character.

==============================
Function extrNumsPlusOne(str As String) As String
Dim re As Object, mc As Object
Set re = CreateObject("vbscript.regexp")
re.Pattern = "\d+[A-Za-z]"
If re.test(str) = True Then
    Set mc = re.Execute(str)
    extrNumsPlusOne = mc(0).Value
End If
End Function
============================
--ron
--ron

Fantastic!  You da man!  This works great!  12800 line items - done!- Hide quoted text -

- Show quoted text -

How do I modify to extract only the number?

In the UDF, just change this line:

re.Pattern = "\d+"

"\d+" tells the function to return the first series of digits in your string.

(The original "\d+[A-Za-z]" told it to return the first series of digits and
the following single letter).

--ron

 

[Rick Rothstein]

Here is a modification to Ron's UDF which will let you specify whether to
return the letter after the number or not...

Function extrNumsPlusOne(str As String, Optional GetChar _
As Boolean = True) As String
Dim re As Object, mc As Object
Set re = CreateObject("vbscript.regexp")
re.Pattern = "\d+"
If GetChar Then re.Pattern = re.Pattern & "[A-Za-z]"
If re.test(str) = True Then
Set mc = re.Execute(str)
extrNumsPlusOne = mc(0).Value
End If
End Function

The default is to return the number and the letter after it; however, if you
specify the optional Boolean argument as False (or 0), only the number is
returned. So, to get only the number, do either of these...

=extrNumsPlusOne(A1,False)
=extrNumsPlusOne(A1,0)

To ge the trailing letter with the number, you can do either of these...

=extrNumsPlusOne(A1)
=extrNumsPlusOne(A1,True)
=extrNumsPlusOne(A1,1)

 

[Rick Rothstein]

Here is a modification to Ron's UDF which will let you specify whether to
return the letter after the number or not...

Function extrNumsPlusOne(str As String, Optional GetChar _
As Boolean = True) As String
Dim re As Object, mc As Object
Set re = CreateObject("vbscript.regexp")
re.Pattern = "\d+"
If GetChar Then re.Pattern = re.Pattern & "[A-Za-z]"
If re.test(str) = True Then
Set mc = re.Execute(str)
extrNumsPlusOne = mc(0).Value
End If
End Function

The default is to return the number and the letter after it; however, if you
specify the optional Boolean argument as False (or 0), only the number is
returned. So, to get only the number, do either of these...

=extrNumsPlusOne(A1,False)
=extrNumsPlusOne(A1,0)

To ge the trailing letter with the number, you can do either of these...

=extrNumsPlusOne(A1)
=extrNumsPlusOne(A1,True)
=extrNumsPlusOne(A1,1)

And, just in case you were interested in pursuing the UDF I posted, here it
is modified to work in the same way I the modification I did to Ron's UDF...

Function GetNumberPlusOne(S As String, Optional GetCharacter _
As Boolean = True) As String
Dim X As Long, Y As Long
For X = 1 To Len(S)
If IsNumeric(Mid(S, X, 1)) Then
For Y = X + 1 To Len(S)
If Not IsNumeric(Mid(S, Y, 1)) Then
GetNumberPlusOne = Mid(S, X, Y - X - GetCharacter)
Exit For
End If
Next
Exit Function
End If
Next
End Function

--
Rick (MVP - Excel)

Repeating my last post:

Easy to do with a UDF:

To enter this <alt-F11> opens the VB Editor. Ensure your project is
highlighted in the Project Explorer window, then Insert/Module and paste
the
code below into the window that opens.

To use this, enter a formula of the type:

The pattern (in the UDF below) will find the first series of digits and
a
following single alpha character.

==============================
Function extrNumsPlusOne(str As String) As String
Dim re As Object, mc As Object
Set re = CreateObject("vbscript.regexp")
re.Pattern = "\d+[A-Za-z]"
If re.test(str) = True Then
Set mc = re.Execute(str)
extrNumsPlusOne = mc(0).Value
End If
End Function
============================
--ron
--ron

Fantastic! You da man! This works great! 12800 line items - done!- Hide
quoted text -

- Show quoted text -

How do I modify to extract only the number?

 

[T. Valko]

While that may be an issue, the immediate problem is that the poster is
posting through Google Groups. Google Groups is notorious for inserting
formatting characters into formulas.

I went to Google Groups, found this thread, copied the formula into a cell
and got the error message.

Google had inserted 2 "-" (dashes) into the formula at these locations:

---LEFT
ROW(1:-999)

 

[Rick Rothstein]

Oh, that's cute! I wasn't aware of this problem with Google Groups. I'm
guessing there is no way to prevent that (other than to tell the poster not
to use Google Groups, that is<g>)?

 

[Ron Rosenfeld]

How do I modify to extract only the number?

It occurs to me that you may want something a bit more general. One way to do
that is to have "pattern" as one of the arguments of the UDF.

That being the case, you could enter the code below but use the following
functions:

To return the digits plus one letter:

=ReMid(A1,"\d+[A-Za-z]")

To return only the digits:

=ReMid(A1,"\d+")

============================
Option Explicit
Function ReMid(str As String, sPattern As String) As String
Dim re As Object, mc As Object
Set re = CreateObject("vbscript.regexp")
re.Pattern = sPattern
If re.test(str) = True Then
Set mc = re.Execute(str)
ReMid = mc(0).Value
End If
End Function
========================
--ron

 

[Ron Rosenfeld]

Sorry, Ron... I mis-posted this to your message instead of the OP's message.

No problem.

After first posting something, I realized he probably wanted an "either/or"
rather than just a change to numbers, so I just posted a more generalized
version which just makes "Pattern" one of the arguments in the UDF. That'd
make further modifications simpler, I think.
--ron

 

[Ron Rosenfeld]

Well, simpler if you know RegEx patterns that is. <g>

That's true. There are similarities to the character lists used for the VBA
Like operator though. There are some shortcuts. For example \d is equivalent
to [0-9]. And there are modifiers so that patterns can be easily repeated.

And I do find something like "/d+[A-Za-z]" quicker to devise as a solution to
the OP's problem than:

=MID(A1,MIN(FIND({0,1,2,3,4,5,6,7,8,9},A1&"0123456789")),
LEN(LOOKUP(9E+307,--LEFT(MID(A1,MIN(FIND(
{0,1,2,3,4,5,6,7,8,9},A1&"0123456789")),999),ROW(1:999))))
+1+FIND(LOOKUP(9E+307,--LEFT(MID(A1,MIN(FIND(
{0,1,2,3,4,5,6,7,8,9},A1&"0123456789")),999),ROW(1:999))),A1)
-MIN(FIND({0,1,2,3,4,5,6,7,8,9},A1&"0123456789")))


--ron

 

[KLZA]

How do I modify to extract only the number?

It occurs to me that you may want something a bit more general.  One way to do
that is to have "pattern" as one of the arguments of the UDF.

That being the case, you could enter the code below but use the following
functions:

To return the digits plus one letter:

=ReMid(A1,"\d+[A-Za-z]")

To return only the digits:

=ReMid(A1,"\d+")

============================
Option Explicit
Function ReMid(str As String, sPattern As String) As String
Dim re As Object, mc As Object
Set re = CreateObject("vbscript.regexp")
re.Pattern = sPattern
If re.test(str) = True Then
    Set mc = re.Execute(str)
    ReMid = mc(0).Value
End If
End Function
========================
--ron

Rick, Ron. Thanks for all of your help. I used the modules and they
work great. I love the support on these groups. Very grreatful.
Thanks again..

 

[Rick Rothstein]

What? It took almost no time to devise that formula. Well, maybe I
exaggerated there just a **wee** bit.<g>

By the way, \d is also equivalent to # also (it is a shortcut for [0-9]) in
a Like operator pattern.

--
Rick (MVP - Excel)

Well, simpler if you know RegEx patterns that is. <g>

That's true. There are similarities to the character lists used for the
VBA
Like operator though. There are some shortcuts. For example \d is
equivalent
to [0-9]. And there are modifiers so that patterns can be easily
repeated.

And I do find something like "/d+[A-Za-z]" quicker to devise as a solution
to
the OP's problem than:

=MID(A1,MIN(FIND({0,1,2,3,4,5,6,7,8,9},A1&"0123456789")),
LEN(LOOKUP(9E+307,--LEFT(MID(A1,MIN(FIND(
{0,1,2,3,4,5,6,7,8,9},A1&"0123456789")),999),ROW(1:999))))
+1+FIND(LOOKUP(9E+307,--LEFT(MID(A1,MIN(FIND(
{0,1,2,3,4,5,6,7,8,9},A1&"0123456789")),999),ROW(1:999))),A1)
-MIN(FIND({0,1,2,3,4,5,6,7,8,9},A1&"0123456789")))


--ron