I'm trying to Replace a Cell with other Cell Text, the think is i need to removed some text from an Image URLs using Excel Formula.
Image URL : domain.com/images/products/63/63/19787/2/279/image-name.jpg
Text that i needed to be removed is 279/. But i need it to be removed at the exact place like domain.com/xxx/xxx/xxx/xxx/xxx/x/279/image-name.jpg between URL and Image name.
I've tried to split it first with this Formula
=MID(A1,FIND("|",SUBSTITUTE(A1,"/","|",LEN(A1)-LEN(SUBSTITUTE(A1,"/",""))-1)),LEN(A1))
Result after i used the Formula
/279/image-name.jpg
and i tried using Replace Formula to replace text from other cell text. Before that i removed the /279 from the result so its only /image-name.jpg now.
=REPLACE(A1, FIND(B1,A1), 4, C1)
But its keep giving me double result in the end of the text like this
domain.com/images/products/63/63/19787/2/image-name.jpg/image-name.jpg
result should be
domain.com/images/products/63/63/19787/2/image-name.jpg - without 279/
is there any problem with the Replace Formula ? or is there any other simpler Formula to make it work ?
To find the location of the next to last slash:
=FIND(CHAR(1),SUBSTITUTE(A1,"/",CHAR(1),LEN(A1)-LEN(SUBSTITUTE(A1,"/",""))-1))
If the contents of that node will always be three characters, you can use replace:
=REPLACE(A1,FIND(CHAR(1),SUBSTITUTE(A1,"/",CHAR(1),LEN(A1)-LEN(SUBSTITUTE(A1,"/",""))-1)),FIND(CHAR(1),SUBSTITUTE(A1,"/",CHAR(1),LEN(A1)-LEN(SUBSTITUTE(A1,"/",""))-1)),"")
If the contents of that node will be a variable number of characters, then we first return the part up to that node, and concatenate with the last node:
=LEFT(A1,FIND(CHAR(1),LEFT(SUBSTITUTE(A1,"/",CHAR(1),LEN(A1)-LEN(SUBSTITUTE(A1,"/",""))-1),99))) &
TRIM(RIGHT(SUBSTITUTE(A1,"/",REPT(" ",99)),99))
EDIT Logic added to ensure that the next to last node is equal to 279
If you need to confirm that the next to last node contains 279, you can check it with:
=ISNUMBER(N(FIND("/279/",MID(A1,FIND(CHAR(1),SUBSTITUTE(A1,"/",CHAR(1),LEN(A1)-LEN(SUBSTITUTE(A1,"/",""))-1)),99))=1))
Using that as part of an IF will return the original string if 279 is not the contents of that node, and replace it only if it is:
=IF(ISNUMBER(N(FIND("/279/",MID(A1,FIND(CHAR(1),SUBSTITUTE(A1,"/",CHAR(1),LEN(A1)-LEN(SUBSTITUTE(A1,"/",""))-1)),99))=1)),
LEFT(A1,FIND(CHAR(1),LEFT(SUBSTITUTE(A1,"/",CHAR(1),LEN(A1)-LEN(SUBSTITUTE(A1,"/",""))-1),99))) &
TRIM(RIGHT(SUBSTITUTE(A1,"/",REPT(" ",99)),99)),A1)
If you happen to have access to TEXTJOIN then use:
=TEXTJOIN("/",,FILTERXML("<t><s>"&SUBSTITUTE(A1,"/","</s><s>")&"</s></t>","//s[not(position() = last()-1)]"))
And if you need to check if it's equal to '279' before removal:
=TEXTJOIN("/",,FILTERXML("<t><s>"&SUBSTITUTE(A1,"/","</s><s>")&"</s></t>","//s[not(position() = last()-1 and .='279')]"))
If you don't have access to TEXTJOIN then you have a fine alternative by #RonRosenfeld.
Another option would be to use REPLACE():
=REPLACE(A1,FIND("|",SUBSTITUTE(A1,"/","|",LEN(A1)-LEN(SUBSTITUTE(A1,"/",""))-1)),FIND("|",SUBSTITUTE(A1,"/","|",LEN(A1)-LEN(SUBSTITUTE(A1,"/",""))))-FIND("|",SUBSTITUTE(A1,"/","|",LEN(A1)-LEN(SUBSTITUTE(A1,"/",""))-1)),"")
You want to use the SUBSTITUTE() function.
The below finds "/279/image-name.jpg" and replaces it with "/image-name.jpg". The added /image-name.jpg will ensure other instances of /279 will remain unaltered.
A1 value = "domain.com/images/products/63/63/19787/2/279/image-name.jpg"
=SUBSTITUTE(A1,"/279/image-name.jpg","/image-name.jpg")
output:
domain.com/images/products/63/63/19787/2/image-name.jpg
Related
Wanted to check if CONCATENATE is the one to use (not sure if my excel has TEXTJOIN), and how to show just the text that has empty value in the cells.
For example in my attachment below, I want the intended result shown like in B2 and B3, where the texts shown with delimiter, when the values are false (empty).
If I were to use CONCATENATE like in Row 10 and Row 11, it's rather manual and it only capture "positive values" as in non-blank cells.
Purpose: To show pending tasks (empty/blank status cells)
Use MID with CONCATENATED IFS:
=MID(IF(C2="","/"&$C$1,"")&IF(D2="","/"&$D$1,"")&IF(E2="","/"&$E$1,"")&IF(F2="","/"&$F$1,"")&IF(GC2="","/"&$G$1,"")&IF(H2="","/"&$H$1,""),2,999)
I would use TEXJOIN and FILTER if you have the newest version of Excel.
For example: =TEXTJOIN("/",1,FILTER($E$2:$I$2, ISBLANK(E3:I3)))
EDIT: For older versions, a temporary workaround is as follows:
make a temporary array the same size as your original dataframe where each value is determined by a formula such as =IF(ISBLANK(E3), E$2&"/","")
Use something like =LEFT(CONCAT(E15:J15), LEN(CONCAT(E15:J15))-1) to get the desired result (where E15:J15 is where I elected to store the first row of the temporary array created in step 1).
I am not sure of your Excel version, but I think this would work in older versions (formatted for readability - will work if you paste it directly into cell B2 and copy down):
=LEFT(CONCAT( INDEX( CHOOSE({1;2;3},$C$1:$H$1,{"/","/","/","/","/","/"},{"","","","","",""}),
INDEX( IF(ISBLANK(C2:H2),{1;2},{3;3}),
MOD(COLUMN(A1:INDEX(1:1,,12))-1,2)+1,
(COLUMN(A1:INDEX(1:1,,12))-1)/2+1 ),
(COLUMN(A1:INDEX(1:1,,12))-1)/2+1 ) ),
SUM(7*ISBLANK(C2:H2))-1 )
Notes
As this is an array formula, you may have to enter it with CTRL + SHIFT + ENTER with an older version of Excel.
The stat labels must all have a length of 6 characters as shown in your post. If not, then they must at least have the same length and the last line SUM(7*ISBLANK(C2:H2))-1 must be changed to replace the 7 with the string length + 1, e.g. a length of 9 would be SUM(10*ISBLANK(C2:H2))-1.
If they don't have the same length, the LEFT( can be removed along with the SUM(10*ISBLANK(C2:H2))-1) at the end. You will end up having a trailing / delimiter at the end. You could fix that for the case of stat F being the last part by changing {"/","/","/","/","/","/"} to {"/","/","/","/","/",""}, but the other cases would still have a trailing /. Another approach is much more complex, but the component SUM(10*ISBLANK(C2:H2))-1) could be shaped to identify what to cut off or maybe a helper column could be built - in any case, let's hope your situation is that the stat labels all have the same length.
The delimiter "/" can be changed, but must always be a single character. If not, then then last line must be changed to SUM( [label length + delimiter length] *ISBLANK(C2:H2))-1.
This formula is fixed to 6 stat columns. If you need for it to accommodate more, it is possible by extending the {"/","/","/","/","/","/"} and {"","","","","",""} (one element for each new column) and replacing every 12 with 2 times the number of columns. Also, obviously, the references $C$1:$H$1 and C1:H2 must be changed to read in your new columns.
I try copy text before last /
Sample:
14501/30305
14501/30305
14507/39756
15936/15943/15954
14507/15940/30405
15936/15943/15980
Result:
14501
14501
14507
15936/15943
14507/15940
15936/15943
I try somelike this but no work in my excel. Return error invalid formula.
=IF(ISERROR(FIND(" ",A2,FIND(" ",A2,1)+1)),A2,LEFT(A2,FIND(" ",A2,FIND(" ",A2,1)+1)))
Everytime I get issue This formula is invalid:
Not a clear answer, but you may go this way (I used cell A1 for my example):
= LEN(A1)-LEN(SUBSTITUTE(A1;"/";"") - This will count how many / you have in A1 cell
Then you may use this count in SUBSTITUTE function:
= SUBSTITUTE(A1;"/";"#";LEN(A1)-LEN(SUBSTITUTE(A1;"/";"")))
in order to replace last / occurrence with some other unique symbol - # in my example
Use FIND to locate this new symbol starting search from 1st symbol:
=FIND("#"; SUBSTITUTE(A1;"/";"#";LEN(A1)-LEN(SUBSTITUTE(A1;"/";""))); 1)
it will give you localtion of #, i.e. location of last / in initial string.
Use this location for LEFT function:
=LEFT(A1; FIND("#"; SUBSTITUTE(A1;"/";"#";LEN(A1)-LEN(SUBSTITUTE(A1;"/";""))); 1)-1)
Just throwing in another version of SUBSTITUTE|RIGHT in the ring as others have already demonstrated.
=SUBSTITUTE(A2,"/"&TRIM(RIGHT(SUBSTITUTE(A2,"/",REPT(" ",99)),99)),"")
You could use multiple ways, for example:
MATCH() the last occurence of the forward slash:
=LEFT(A1,MATCH(2,1/(MID(A1,SEQUENCE(LEN(A1)),1)="/"))-1)
Or if one does not have Microsoft365:
=LEFT(A1,MATCH(2,1/(MID(A1,ROW(A1:INDEX(A:A,LEN(A1))),1)="/"))-1)
SUBSTITUTE() the last occurence of the forward slash:
=LEFT(A1,FIND("|",SUBSTITUTE(A1,"/","|",LEN(A1)-LEN(SUBSTITUTE(A1,"/",""))))-1)
TEXTJOIN() any other than last part:
=TEXTJOIN("/",,FILTERXML("<t><s>"&SUBSTITUTE(A1,"/","</s><s>")&"</s></t>","//s[position()<last()]"))
Edit:
If this is done in GS for now, you can use REGEXEXTRACT() inside and INDEX():
=INDEX(IF(A2:A="",,REGEXEXTRACT(A2:A,"(^.*)\/")),,)
This will spill all your answers down in the same column.
I have Excel sheet which contains data similar to
Addresses
xyz,abc,olk
opn,opk,prt
we-ylj,tyf,uyfas
oiui,ytfy,tydry - We also work in bla,bla,bla
ytfyt,tyfyt,ghfyt
i-hgsd,gsdf-hgd,sdgh,- We also work in xxx,yy,zzz
ytsfgh,gfasdg,tydsfyt
I want to remove all substring which is next to the character "-" only if it's in the last position.
Result should be like
xyz,abc,olk
opn,opk,prt
we-ylj,tyf,uyfas
oiui,ytfy,tydry
ytfyt,tyfyt,ghfyt i-hgsd,gsdf-hgd,sdgh
ytsfgh,gfasdg,tydsfyt
I tried with =Substitute function but unable to replace data because of the last substring separated from "-" is not similar.
Going by your specifications, I would use two columns just so it's not a very long formula:
In B1:
=IFERROR(FIND(CHAR(1),SUBSTITUTE(A1,"-",CHAR(1),LEN(A1)-LEN(SUBSTITUTE(A1,"-",""))))-1,LEN(A1))
This gets the position of the last - or the full text length.
Then in C1:
=LEFT(A1,IF(FIND(",",A1)<B1,B1,LEN(A1)))
This checks if there's a , before the last -. If there is no ,, then the full text is taken.
EDIT: I only now noticed your edited comment. If it's just everything after - We, then I would use this:
=TRIM(LEFT(A1,IFERROR(FIND("- We",A1)-2,LEN(A1))))
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 /
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 /