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How can I find an entire number between a "#" and a space when that combination could appear anywhere in a given cell?
Example cell contents:
"This is a #123 Test that I 45 like to run"
"This is a #45 Test that I 98 like to run"
I need to return "123" from the first one and "45" from the second one.
Using Mid(), I can return the "1", but the problem is the number between # and space can vary in length, but there will generally be a #, number or numbers, then a space.
As a secondary issue, there may be scenarios where there is no "#", but I need to find the first numeric value in the cell and return them (i.e. "1", "34", "648").
Any advice on either of these challenges is greatly appreciated.
This should work as well:
=MID(A11,(FIND("#",A11,1)+1),FIND(" ",A11,FIND("#",A11,1)+1)-FIND("#",A11,1))
works by looking for the hash and the following space... Not for the secondary question...
Since you've put the excel-vba tag on your question, here's a vba way of doing it using regular expressions that should satisfy both your primary and secondary issues:
Sub tmp()
Dim regEx As New RegExp
regEx.Pattern = "^.*?\#?(\d+)"
Dim i As Integer
For i = 1 To Range("A" & Rows.Count).End(xlUp).Row:
Set mat = regEx.Execute(Cells(i, 1).Value)
If mat.Count = 1 Then
Cells(i, 2).Value = mat(0).SubMatches(0)
End If
Next
End Sub
The regular expression uses a non-greedy character search (ie the "?" on the end of "'.*?" is what does that) to find the first pattern in the cell that matches either "#123" or just "123" where the "123" is any arbitrary sequence of digits.
This will return the first number in a string:
=--LEFT(MID(A1,AGGREGATE(15,6,FIND({1,2,3,4,5,6,7,8,9,0},A1),1),LEN(A1)),FIND(" ",MID(A1,AGGREGATE(15,6,FIND({1,2,3,4,5,6,7,8,9,0},A1),1),LEN(A1))))
AGGREGATE was introduced in 2010 Excel. If you do not ahve that then you will need to use this array formula:
=--LEFT(MID(A1,MIN(IFERROR(FIND({1,2,3,4,5,6,7,8,9,0},A1),1E+99)),LEN(A1)),FIND(" ",MID(A1,MIN(IFERROR(FIND({1,2,3,4,5,6,7,8,9,0},A1),1E+99)),LEN(A1))))
Being an array formula it needs to be confirmed with Ctrl-Shift-Enter instead of Enter when exiting edit mode. If done correctly then excel will put {} around the formula.
I have here some text strings
"16cg-301 -request","16cg-3368 - for review","16cg-3684 - for process"
what i would like to do is to remove all the text and characters except the number and the letters "cg" and - which is within the reference code.
If the string you want to extract is always before the first space in the full string then you can use SEARCH and LEFT to extract your reference code:
=LEFT(A1,SEARCH(" ",A1)-1)
This formula would take 16cg-3368 from 16cg-3368 - for review.
I suggest using something like suggested here
How to use Regular Expressions (Regex) in Microsoft Excel both in-cell and loops
With a replace regex similar to this
[^\dcg]*
or a match regex like this
^([0-9cg- ]+).*
else you could also work with a strange formule similar to this
=CONCATENATE(IF(NOT(ISERROR(SEARCH(MID(A2;1;1);"01234567890cg-")>0));MID(A2;1;1);"");IF(NOT(ISERROR(SEARCH(MID(A2;2;1);"01234567890cg-")>0));MID(A2;2;1);"");IF(NOT(ISERROR(SEARCH(MID(A2;3;1);"01234567890cg-")>0));MID(A2;3;1);"");IF(NOT(ISERROR(SEARCH(MID(A2;4;1);"01234567890cg-")>0));MID(A2;4;1);"");IF(NOT(ISERROR(SEARCH(MID(A2;5;1);"01234567890cg-")>0));MID(A2;5;1);"");IF(NOT(ISERROR(SEARCH(MID(A2;6;1);"01234567890cg-")>0));MID(A2;6;1);"");IF(NOT(ISERROR(SEARCH(MID(A2;7;1);"01234567890cg-")>0));MID(A2;7;1);"");IF(NOT(ISERROR(SEARCH(MID(A2;8;1);"01234567890cg-")>0));MID(A2;8;1);"");IF(NOT(ISERROR(SEARCH(MID(A2;9;1);"01234567890cg-")>0));MID(A2;9;1);"");IF(NOT(ISERROR(SEARCH(MID(A2;10;1);"01234567890cg-")>0));MID(A2;10;1);"");IF(NOT(ISERROR(SEARCH(MID(A2;11;1);"01234567890cg-")>0));MID(A2;11;1);"");IF(NOT(ISERROR(SEARCH(MID(A2;12;1);"01234567890cg-")>0));MID(A2;12;1);"");IF(NOT(ISERROR(SEARCH(MID(A2;13;1);"01234567890cg-")>0));MID(A2;13;1);"");IF(NOT(ISERROR(SEARCH(MID(A2;14;1);"01234567890cg-")>0));MID(A2;14;1);"");IF(NOT(ISERROR(SEARCH(MID(A2;15;1);"01234567890cg-")>0));MID(A2;15;1);"");IF(NOT(ISERROR(SEARCH(MID(A2;16;1);"01234567890cg-")>0));MID(A2;16;1);"");IF(NOT(ISERROR(SEARCH(MID(A2;17;1);"01234567890cg-")>0));MID(A2;17;1);"");IF(NOT(ISERROR(SEARCH(MID(A2;18;1);"01234567890cg-")>0));MID(A2;18;1);"");IF(NOT(ISERROR(SEARCH(MID(A2;19;1);"01234567890cg-")>0));MID(A2;19;1);"");IF(NOT(ISERROR(SEARCH(MID(A2;20;1);"01234567890cg-")>0));MID(A2;20;1);"");IF(NOT(ISERROR(SEARCH(MID(A2;21;1);"01234567890cg-")>0));MID(A2;21;1);"");IF(NOT(ISERROR(SEARCH(MID(A2;22;1);"01234567890cg-")>0));MID(A2;22;1);"");IF(NOT(ISERROR(SEARCH(MID(A2;23;1);"01234567890cg-")>0));MID(A2;23;1);"");IF(NOT(ISERROR(SEARCH(MID(A2;24;1);"01234567890cg-")>0));MID(A2;24;1);"");IF(NOT(ISERROR(SEARCH(MID(A2;25;1);"01234567890cg-")>0));MID(A2;25;1);"");IF(NOT(ISERROR(SEARCH(MID(A2;26;1);"01234567890cg-")>0));MID(A2;26;1);"");IF(NOT(ISERROR(SEARCH(MID(A2;27;1);"01234567890cg-")>0));MID(A2;27;1);"");IF(NOT(ISERROR(SEARCH(MID(A2;28;1);"01234567890cg-")>0));MID(A2;28;1);"");IF(NOT(ISERROR(SEARCH(MID(A2;29;1);"01234567890cg-")>0));MID(A2;29;1);"");IF(NOT(ISERROR(SEARCH(MID(A2;30;1);"01234567890cg-")>0));MID(A2;30;1);"");IF(NOT(ISERROR(SEARCH(MID(A2;31;1);"01234567890cg-")>0));MID(A2;31;1);"");IF(NOT(ISERROR(SEARCH(MID(A2;32;1);"01234567890cg-")>0));MID(A2;32;1);""))
only works by now for less than 33 signs.
problem here will be that you will get unexpected behavior like this:
123cg-123 - Process => 123cg-123-c
after rereading , I think you should try an other approach than described in the question ;-)
If you want to return everything up to and including the last digit, then try:
=LEFT(A1,LOOKUP(2,1/ISNUMBER(-MID(A1,seq,1)),seq))
seq is a named formula: Formula ► Define Name
Name: seq
Refers to: =ROW(INDEX($1:$65535,1,1):INDEX($1:$65535,255,1))
seq returns an array of sequential numbers from 1 to 255.
mid(a1,seq,1)
returns an array consisting of the individual characters in the string in A1. The leading minus sign converts the digits from strings to numbers.
The lookup function will then return the position of the last digit
Is there an efficient way to identify the last character/string match in a string using base functions? I.e. not the last character/string of the string, but the position of a character/string's last occurrence in a string. Search and find both work left-to-right so I can't think how to apply without lengthy recursive algorithm. And this solution now seems obsolete.
I think I get what you mean. Let's say for example you want the right-most \ in the following string (which is stored in cell A1):
Drive:\Folder\SubFolder\Filename.ext
To get the position of the last \, you would use this formula:
=FIND("#",SUBSTITUTE(A1,"\","#",(LEN(A1)-LEN(SUBSTITUTE(A1,"\","")))/LEN("\")))
That tells us the right-most \ is at character 24. It does this by looking for "#" and substituting the very last "\" with an "#". It determines the last one by using
(len(string)-len(substitute(string, substring, "")))\len(substring)
In this scenario, the substring is simply "\" which has a length of 1, so you could leave off the division at the end and just use:
=FIND("#",SUBSTITUTE(A1,"\","#",LEN(A1)-LEN(SUBSTITUTE(A1,"\",""))))
Now we can use that to get the folder path:
=LEFT(A1,FIND("#",SUBSTITUTE(A1,"\","#",LEN(A1)-LEN(SUBSTITUTE(A1,"\","")))))
Here's the folder path without the trailing \
=LEFT(A1,FIND("#",SUBSTITUTE(A1,"\","#",LEN(A1)-LEN(SUBSTITUTE(A1,"\",""))))-1)
And to get just the filename:
=MID(A1,FIND("#",SUBSTITUTE(A1,"\","#",LEN(A1)-LEN(SUBSTITUTE(A1,"\",""))))+1,LEN(A1))
However, here is an alternate version of getting everything to the right of the last instance of a specific character. So using our same example, this would also return the file name:
=TRIM(RIGHT(SUBSTITUTE(A1,"\",REPT(" ",LEN(A1))),LEN(A1)))
How about creating a custom function and using that in your formula? VBA has a built-in function, InStrRev, that does exactly what you're looking for.
Put this in a new module:
Function RSearch(str As String, find As String)
RSearch = InStrRev(str, find)
End Function
And your function will look like this (assuming the original string is in B1):
=LEFT(B1,RSearch(B1,"\"))
New Answer | 31-3-2022:
With even newer functions come even shorter answers. At time of writing in BETA, but probably widely available in the near future, we can use TEXTBEFORE():
=LEN(TEXTBEFORE(A2,B2,-1))+1
The trick here is that the 3rd parameter tells the function to retrieve the last occurence of the substring we give in the 2nd parameter. At time of writing this function is still case-sensitive by default which could be handeld by the optional 4th parameter.
Original Answer | 17-6-2020:
With newer versions of excel come new functions and thus new methods. Though it's replicable in older versions (yet I have not seen it before), when one has Excel O365 one can use:
=MATCH(2,1/(MID(A1,SEQUENCE(LEN(A1)),1)="Y"))
This can also be used to retrieve the last position of (overlapping) substrings:
=MATCH(2,1/(MID(A1,SEQUENCE(LEN(A1)),2)="YY"))
| Value | Pattern | Formula | Position |
|--------|---------|------------------------------------------------|----------|
| XYYZ | Y | =MATCH(2,1/(MID(A2,SEQUENCE(LEN(A2)),1)="Y")) | 3 |
| XYYYZ | YY | =MATCH(2,1/(MID(A3,SEQUENCE(LEN(A3)),2)="YY")) | 3 |
| XYYYYZ | YY | =MATCH(2,1/(MID(A4,SEQUENCE(LEN(A4)),2)="YY")) | 4 |
Whilst this both allows us to no longer use an arbitrary replacement character and it allows overlapping patterns, the "downside" is the useage of an array.
Note: You can force the same behaviour in older Excel versions through either
=MATCH(2,1/(MID(A2,ROW(A1:INDEX(A:A,LEN(A2))),1)="Y"))
Entered through CtrlShiftEnter, or using an inline INDEX to get rid of implicit intersection:
=MATCH(2,INDEX(1/(MID(A2,ROW(A1:INDEX(A:A,LEN(A2))),1)="Y"),))
tigeravatar and Jean-François Corbett suggested to use this formula to generate the string right of the last occurrence of the "\" character
=TRIM(RIGHT(SUBSTITUTE(A1,"\",REPT(" ",LEN(A1))),LEN(A1)))
If the character used as separator is space, " ", then the formula has to be changed to:
=SUBSTITUTE(RIGHT(SUBSTITUTE(A1," ",REPT("{",LEN(A1))),LEN(A1)),"{","")
No need to mention, the "{" character can be replaced with any character that would not "normally" occur in the text to process.
Just came up with this solution, no VBA needed;
Find the last occurance of "_" in my example;
=IFERROR(FIND(CHAR(1);SUBSTITUTE(A1;"_";CHAR(1);LEN(A1)-LEN(SUBSTITUTE(A1;"_";"")));0)
Explained inside out;
SUBSTITUTE(A1;"_";"") => replace "_" by spaces
LEN( *above* ) => count the chars
LEN(A1)- *above* => indicates amount of chars replaced (= occurrences of "_")
SUBSTITUTE(A1;"_";CHAR(1); *above* ) => replace the Nth occurence of "_" by CHAR(1) (Nth = amount of chars replaced = the last one)
FIND(CHAR(1); *above* ) => Find the CHAR(1), being the last (replaced) occurance of "_" in our case
IFERROR( *above* ;"0") => in case no chars were found, return "0"
Hope this was helpful.
You could use this function I created to find the last instance of a string within a string.
Sure the accepted Excel formula works, but it's much too difficult to read and use. At some point you have to break out into smaller chunks so it's maintainable. My function below is readable, but that's irrelevant because you call it in a formula using named parameters. This makes using it simple.
Public Function FindLastCharOccurence(fromText As String, searchChar As String) As Integer
Dim lastOccur As Integer
lastOccur = -1
Dim i As Integer
i = 0
For i = Len(fromText) To 1 Step -1
If Mid(fromText, i, 1) = searchChar Then
lastOccur = i
Exit For
End If
Next i
FindLastCharOccurence = lastOccur
End Function
I use it like this:
=RIGHT(A2, LEN(A2) - FindLastCharOccurence(A2, "\"))
Considering a part of a Comment made by #SSilk my end goal has really been to get everything to the right of that last occurence an alternative approach with a very simple formula is to copy a column (say A) of strings and on the copy (say ColumnB) apply Find and Replace. For instance taking the example: Drive:\Folder\SubFolder\Filename.ext
This returns what remains (here Filename.ext) after the last instance of whatever character is chosen (here \) which is sometimes the objective anyway and facilitates finding the position of the last such character with a short formula such as:
=FIND(B1,A1)-1
I'm a little late to the party, but maybe this could help. The link in the question had a similar formula, but mine uses the IF() statement to get rid of errors.
If you're not afraid of Ctrl+Shift+Enter, you can do pretty well with an array formula.
String (in cell A1):
"one.two.three.four"
Formula:
{=MAX(IF(MID(A1,ROW($1:$99),1)=".",ROW($1:$99)))} use Ctrl+Shift+Enter
Result: 14
First,
ROW($1:$99)
returns an array of integers from 1 to 99: {1,2,3,4,...,98,99}.
Next,
MID(A1,ROW($1:$99),1)
returns an array of 1-length strings found in the target string, then returns blank strings after the length of the target string is reached: {"o","n","e",".",..."u","r","","",""...}
Next,
IF(MID(I16,ROW($1:$99),1)=".",ROW($1:$99))
compares each item in the array to the string "." and returns either the index of the character in the string or FALSE: {FALSE,FALSE,FALSE,4,FALSE,FALSE,FALSE,8,FALSE,FALSE,FALSE,FALSE,FALSE,14,FALSE,FALSE.....}
Last,
=MAX(IF(MID(I16,ROW($1:$99),1)=".",ROW($1:$99)))
returns the maximum value of the array: 14
Advantages of this formula is that it is short, relatively easy to understand, and doesn't require any unique characters.
Disadvantages are the required use of Ctrl+Shift+Enter and the limitation on string length. This can be worked around with a variation shown below, but that variation uses the OFFSET() function which is a volatile (read: slow) function.
Not sure what the speed of this formula is vs. others.
Variations:
=MAX((MID(A1,ROW(OFFSET($A$1,,,LEN(A1))),1)=".")*ROW(OFFSET($A$1,,,LEN(A1)))) works the same way, but you don't have to worry about the length of the string
=SMALL(IF(MID(A1,ROW($1:$99),1)=".",ROW($1:$99)),2) determines the 2nd occurrence of the match
=LARGE(IF(MID(A1,ROW($1:$99),1)=".",ROW($1:$99)),2) determines the 2nd-to-last occurrence of the match
=MAX(IF(MID(I16,ROW($1:$99),2)=".t",ROW($1:$99))) matches a 2-character string **Make sure you change the last argument of the MID() function to the number of characters in the string you wish to match!
In newer versions of Excel (2013 and up) flash fill might be a simple and quick solution see: Using Flash Fill in Excel
.
For a string in A1 and substring in B1, use:
=XMATCH(B1,MID(A1,SEQUENCE(LEN(A1)),LEN(B1)),,-1)
Working from inside out, MID(A1,SEQUENCE(LEN(A1)),LEN(B1)) splits string A1 into a dynamic array of substrings, each the length of B1. To find the position of the last occurrence of substring B1, we use XMATCH with its Search_mode argument set to -1.
A simple way to do that in VBA is:
YourText = "c:\excel\text.txt"
xString = Mid(YourText, 2 + Len(YourText) - InStr(StrReverse(YourText), "\" ))
Very late to the party, but A simple solution is using VBA to create a custom function.
Add the function to VBA in the WorkBook, Worksheet, or a VBA Module
Function LastSegment(S, C)
LastSegment = Right(S, Len(S) - InStrRev(S, C))
End Function
Then the cell formula
=lastsegment(B1,"/")
in a cell and the string to be searched in cell B1 will populate the cell with the text trailing the last "/" from cell B1.
No length limit, no obscure formulas.
Only downside I can think is the need for a macro-enabled workbook.
Any user VBA Function can be called this way to return a value to a cell formula, including as a parameter to a builtin Excel function.
If you are going to use the function heavily you'll want to check for the case when the character is not in the string, then string is blank, etc.
If you're only looking for the position of the last instance of character "~" then
=len(substitute(String,"~",""))+1
I'm sure there is version that will work with the last instance of a string but I have to get back to work.
Cell A1 = find/the/position/of/the last slash
simple way to do it is reverse the text and then find the first slash as normal. Now you can get the length of the full text minus this number.
Like so:
=LEN(A1)-FIND("/",REVERSETEXT(A1),1)+1
This returns 21, the position of the last /
Hello I have a column with strings (names of products) in it.
Now these are formatted as Name LenghtxWidth, example Green box 20x30. Now I need to change the 20 with the 30 in this example so I get Green box 30x20, any ideas how I can achieve this?
Thanks
Here is both a formula solution, as well as a VBA solution using Regular Expressions:
Formula
=LEFT(A1,FIND(TRIM(RIGHT(SUBSTITUTE(A1," ",REPT(" ",99)),99)),A1)-1)&
MID(TRIM(RIGHT(SUBSTITUTE(A1," ",REPT(" ",99)),99)),SEARCH("x",TRIM(RIGHT(SUBSTITUTE(A1," ",REPT(" ",99)),99)))+1,99)&
"x"&
LEFT(TRIM(RIGHT(SUBSTITUTE(A1," ",REPT(" ",99)),99)),SEARCH("x",TRIM(RIGHT(SUBSTITUTE(A1," ",REPT(" ",99)),99)))-1)
UDF
Option Explicit
Function RevWL(S As String)
Dim RE As Object
Const sPat As String = "(\d+.?\d*)x(\d+.?\d*)"
'If L or W might start with a decimal point, and not a digit,
'Then change sPat to: (\d*.?\d+)x(\d*.?\d+)
Set RE = CreateObject("vbscript.regexp")
With RE
.Global = True
.ignorecase = True
.Pattern = sPat
RevWL = .Replace(S, "$2x$1")
End With
End Function
Here is an example of the kinds of data I tested with:
The Formula works by looking at the last space-separated substring which would be LxW, then reversing the portion after and before the x, then concatenating everything back together.
The regex pattern captures the two numbers (could be integers or decimals, so long as the start with an integer -- although that could be changed if needed), and reversing them.
Here is a more detailed explanation of the regex (and the replacement string) with links to a tutorial:
(\d+.?\d*)x(\d+.?\d*)
(\d+.?\d*)x(\d+.?\d*)
Options: Case insensitive; ^$ don’t match at line breaks
Match the regex below and capture its match into backreference number 1 (\d+.?\d*)
Match a single character that is a “digit” \d+
Between one and unlimited times, as many times as possible, giving back as needed (greedy) +
Match any single character that is NOT a line break character .?
Between zero and one times, as many times as possible, giving back as needed (greedy) ?
Match a single character that is a “digit” \d*
Between zero and unlimited times, as many times as possible, giving back as needed (greedy) *
Match the character “x” literally x
Match the regex below and capture its match into backreference number 2 (\d+.?\d*)
Match a single character that is a “digit” \d+
Between one and unlimited times, as many times as possible, giving back as needed (greedy) +
Match any single character that is NOT a line break character .?
Between zero and one times, as many times as possible, giving back as needed (greedy) ?
Match a single character that is a “digit” \d*
Between zero and unlimited times, as many times as possible, giving back as needed (greedy) *
$2x$1
Insert the text that was last matched by capturing group number 2 $2
Insert the character “x” literally x
Insert the text that was last matched by capturing group number 1 $1
Created with RegexBuddy
Here is a VBA solution that will work for you:
Option Explicit
Function Switch(r As Range) As String
Dim measurement As String
Dim firstPart As String
Dim secondPart As String
measurement = Right(r, Len(r) - InStrRev(r, " "))
secondPart = Right(measurement, Len(measurement) - InStr(1, measurement, "x"))
firstPart = Left(measurement, InStr(1, measurement, "x") - 1)
Switch = Left(r, InStrRev(r, " ") - 1) & " " & secondPart & "x" & firstPart
End Function
You can paste this in a regular module in the VBE (Visual Basic Editor) and use it as a regular function/formula. If your value is in cell A1 then type =Switch(A1) in cell B1. Hope it helps!
Ok, so it is really easier to use VBA, but if you want only some formulas you can use some columns to split your text and then concatenate your cells.
Here is a little example:
Of course B1-4 are optional. It is here only to have something more readable, but you can do use only one formula
=CONCATENATE(LEFT(A1, SEARCH(" ",A1,1)-1)," ",RIGHT(RIGHT(A1,LEN(A1)-SEARCH(" ",A1,1)),LEN(RIGHT(A1,LEN(A1)-SEARCH(" ",A1,1)))-SEARCH("x",RIGHT(A1,LEN(A1)-SEARCH(" ",A1,1)),1)),"x",LEFT(RIGHT(A1,LEN(A1)-SEARCH(" ",A1,1)), SEARCH("x",RIGHT(A1,LEN(A1)-SEARCH(" ",A1,1)),1)-1))
If you have several spaces in your names, you can use this formula that will search the last space in the text
=CONCATENATE(LEFT(A1, SEARCH("^^",SUBSTITUTE(A1," ","^^",LEN(A1)-LEN(SUBSTITUTE(A1," ",""))))-1)," ",RIGHT(RIGHT(A1,LEN(A1)-SEARCH("^^",SUBSTITUTE(A1," ","^^",LEN(A1)-LEN(SUBSTITUTE(A1," ",""))))),LEN(RIGHT(A1,LEN(A1)-SEARCH("^^",SUBSTITUTE(A1," ","^^",LEN(A1)-LEN(SUBSTITUTE(A1," ",""))))))-SEARCH("x",RIGHT(A1,LEN(A1)-SEARCH("^^",SUBSTITUTE(A1," ","^^",LEN(A1)-LEN(SUBSTITUTE(A1," ",""))))),1)),"x",LEFT(RIGHT(A1,LEN(A1)-SEARCH("^^",SUBSTITUTE(A1," ","^^",LEN(A1)-LEN(SUBSTITUTE(A1," ",""))))), SEARCH("x",RIGHT(A1,LEN(A1)-SEARCH("^^",SUBSTITUTE(A1," ","^^",LEN(A1)-LEN(SUBSTITUTE(A1," ",""))))),1)-1))
I am interested in removing leading alphabetical (alpha) characters from cells which appear in a column. I only wish to remove the leading alpha characters (including UPPER and LOWER case): if alpha characters appear after a number they should be kept. Some cells in the column might not have leading alpha characters.
Here is an example of what I have:
36173
PIL51014
4UNV22001
ZEB54010
BICMPAFG11BK
BICMPF11
Notice how there are not always the same number of leading alpha characters. I cannot simply use a Left or Right function in Excel, because the number of characters I wish to keep and remove varies.
A correct output for what I am looking for would look like:
36173
51014
4UNV22001
54010
11BK
11
Notice how the second to last row preserved the characters "BK", and the 3rd row preserved "UNV". I cannot simply remove all alpha characters.
I am a beginner with visual basic and was not able to figure out how to use excel functions to address my issue. How would I do this?
Here is an Excel formula that will "strip off the leading alpha characters" Actually, it looks for the first numeric character, and returns everything after that:
=MID(A1,MIN(FIND({0;1;2;3;4;5;6;7;8;9},A1&"0123456789")),99)
The 99 at the end needs to be some value longer than the longest string you might be processing. 99 usually works.
Here's a formula based solution complete with test results:
=MID(A1,MIN(SEARCH({0,1,2,3,4,5,6,7,8,9},A1&"0123456789"),255),100)
Change the 100 at the end if any string may be longer than 100 characters. Also the 255 is not needed, but it won't hurt.
This short UDF should strip off leading alphabetic characters.
Function noLeadAlpha(str As String)
If Not IsNumeric(str) Then
Do While Asc(str) < 48 Or Asc(str) > 57
str = Mid(str, 2)
If Not CBool(Len(str)) Then Exit Do
Loop
End If
noLeadAlpha = str
End Function
Koodos Jeeped, you beat me to it.
But here is an alternative anyway:
Function RemoveAlpha(aString As String) As String
For i = 1 To Len(aString)
Select Case Mid(aString, i, 1)
Case "0" To "9"
RemoveAlpha = Right(aString, Len(aString) - i + 1): Exit For
End Select
Next i
End Function