Excel: last character/string match in a string

问题:

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.

回答1:

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))) 

回答2:

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,"\")) 

回答3:

=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.

回答4:

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, "\")) 

回答5:

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.

回答6:

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! 

回答7:

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 

回答8:

A simple way to do that in VBA is:

YourText = "c:\excel\text.txt" xString = Mid(YourText, 2 + Len(YourText) - InStr(StrReverse(YourText), "\" )) 

回答9:

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.

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