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I am not sure where to begin with the formula as I have gotten myself so confused with everything. I have a cell the contains "PON " or "PON: " or "PON = " then the actual PON (Example: PON 123467) I want to formula to return 123467 in the cell.
Examples What I want returned
I have PON 123467 for shoes 123467
I have PON: 234567-AB for food 234567-AB
I have PON - 569874-Weird for accessories 569874-Weird
I have PON = DOG-564-987 for dog food DOG-564-987
I am currently using Excel 365
Filterxml() will give you best companion here in this case. Try-
=FILTERXML("<t><s>"&SUBSTITUTE(FILTERXML("<t><s>"&SUBSTITUTE(A1," for","</s><s>")&"</s></t>","//s[1]")," ","</s><s>")&"</s></t>","//s[last()]")
Using FILTERXML, and testing for a substring following PON, you can try:
=FILTERXML("<t><s>"&SUBSTITUTE(TRIM(A1)," ","</s><s>") & "</s></t>","//s[contains(.,'PON')]/following-sibling::*[string-length(.)>2][1]")
Note that FILTERXML solution will cause a PON that is solely numeric, but with a leading zero, to drop the leading zero. Unfortunately, the xPath implementation in that function does not include the string() function
If dropping the leading zero might be a problem, you can add a character to the node that will force the number to be seen as a string. In the modified formula below, I use the unicode zero-width space, but there are others you can use. Note that this will count as a character for the string=length function, so be sure to maintain the >2 parameter:
=FILTERXML("<t><s>"&SUBSTITUTE(TRIM(A1)," ","</s><s>"&UNICHAR(8203)) & "</s></t>","//s[contains(.,'PON')]/following-sibling::*[string-length(.)>2][1]")
Because of the variablity in your data, that sometimes there are extraneous space-separated substrings between PON and your desired extract, the xpath:
locates the substring PON
returns all subsequent siblings that have a string-length of more than two (adjust if necessary)
returns the first sibling that meets that criterion.
You might try this formula.
=TRIM(LEFT(MID(A2,FIND(#{1,2,3,4,5,6,7,8,9},A2),100),FIND(" ",MID(A2,FIND(#{1,2,3,4,5,6,7,8,9},A2),100))))
It extracts the text between the first number and the first space following that number. The size of that extract is limited to 100 characters.
I want to change so many dates saved as text in excel sheets to Arabic (Hijri) date format, ie; dd/mm/yyyy to (right to left) yyyy/mm/dd .. it can be done manually be change custom format from Number Format panel ( by choosing the location , calendar type and the right format from list ) then replace the text in cell by same value ..
I need VBA code to automate this replacement process for any ltr date and for existing rtl date just convert the format from general to date ..
another problem , there is some other text around the date in cell like ( dd/mm/yyyy ttt ) . I want the code to remove this text (ttt) ( any text ) and then change to the right format
I found this code but it is not work to my specific need
Changing the date format to yyyy-mm-dd
I appreciate any help , thanks in advance ..
You have a number of separate problems and it is unrealistic to expect someone else to have posted a complete solution to that set of problems or that someone will code a complete solution for you. You need to split your total problem into its components and create or look for a solution to each component.
You have strings that contain CE dates in the format “dd/mm/yyyy”. These dates could be surrounded by text. You give the example “dd/mm/yyyy ttt”. Can ttt contain spaces? Could the “ttt” come before the date? Could the string be as complicated as “aaaa bbbb cccc 12/11/2016 dddd eeee ffff”?
Whatever the situation, I suspect something like:
Dim Part() As String
Part = Split(.Cells(R, C).Value," ")
would be the core of first step. With my complicated example, this would give:
Part(0) = "aaaa"
Part(1) = "bbbb"
Part(2) = "cccc"
Part(3) = "12/11/2016"
: : : :
A loop over the parts of each cell value looking for a string for which IsDate gives True would allow you to find the date so .Cells(R, C).Value = Part(N) would delete the unwanted text.
I would take a copy of your data and try to code a macro that discards the unwanted text. If you can successfully create that macro, you have completed step 1 of your solution. If you have trouble with this macro, you can ask for help here and expect to get it.
The next step is to convert the string “dd/mm/yyyy” to an Excel date. Excel holds dates as the number of dates since 1/1/1900 CE. Replacing:
.Cells(R, C).Value = Part(N)by .Cells(R, C).Value = CDate(Part(N))ought to do the trick. However, Excel sometimes tries to interpret “dd/mm/yyyy” dates as “mm/dd/yyyy”. I think you will be alright but be aware of this possibility.
Your last step is to convert a date from the CE calendar to the Hijri calendar. This is not just a format issue. The two calendars have different year zeroes and different month lengths. There may be a standard conversion function in your country but there does not appear to be one here in the UK. There is help online so you should be able to find a function that will perform the conversion.
You have a number of separate problems and it is unrealistic to expect someone else to have posted a complete solution to that set of problems or that someone will code a complete solution for you. You need to split your total problem into its components and create or look for a solution to each component.
You have strings that contain CE dates in the format “dd/mm/yyyy”. These dates could be surrounded by text. You give the example “dd/mm/yyyy ttt”. Can ttt contain spaces? Could the “ttt” come before the date? Could the string be as complicated as “aaaa bbbb cccc 12/11/2016 dddd eeee ffff”?
Whatever the situation, I suspect something like:
Dim Part() As String
Part = Split(.Cells(R, C).Value," ")
would be the core of first step. With my complicated example, this would give:
Part(0) = "aaaa"
Part(1) = "bbbb"
Part(2) = "cccc"
Part(3) = "12/11/2016"
: : : :
A loop over the parts of each cell value looking for a string for which IsDate gives True would allow you to find the date so .Cells(R, C).Value = Part(N) would delete the unwanted text.
I would take a copy of your data and try to code a macro that discards the unwanted text. If you can successfully create that macro, you have completed step 1 of your solution. If you have trouble with this macro, you can ask for help here and expect to get it.
The next step is to convert the string “dd/mm/yyyy” to an Excel date. Excel holds dates as the number of dates since 1/1/1900 CE. Replacing:
.Cells(R, C).Value = Part(N)by .Cells(R, C).Value = CDate(Part(N))ought to do the trick. However, Excel sometimes tries to interpret “dd/mm/yyyy” dates as “mm/dd/yyyy”. I think you will be alright but be aware of this possibility.
Your last step is to convert a date from the CE calendar to the Hijri calendar. This is not just a format issue. The two calendars have different year zeroes and different month lengths. There may be a standard conversion function in your country but there does not appear to be one here in the UK. There is help online so you should be able to find a function that will perform the conversion.
You have a number of separate problems and it is unrealistic to expect someone else to have posted a complete solution to that set of problems or that someone will code a complete solution for you. You need to split your total problem into its components and create or look for a solution to each component.
You have strings that contain CE dates in the format “dd/mm/yyyy”. These dates could be surrounded by text. You give the example “dd/mm/yyyy ttt”. Can ttt contain spaces? Could the “ttt” come before the date? Could the string be as complicated as “aaaa bbbb cccc 12/11/2016 dddd eeee ffff”?
Whatever the situation, I suspect something like:
Dim Part() As String
Part = Split(.Cells(R, C).Value," ")
would be the core of first step. With my complicated example, this would give:
Part(0) = "aaaa"
Part(1) = "bbbb"
Part(2) = "cccc"
Part(3) = "12/11/2016"
: : : :
A loop over the parts of each cell value looking for a string for which IsDate gives True would allow you to find the date so .Cells(R, C).Value = Part(N) would delete the unwanted text.
I would take a copy of your data and try to code a macro that discards the unwanted text. If you can successfully create that macro, you have completed step 1 of your solution. If you have trouble with this macro, you can ask for help here and expect to get it.
The next step is to convert the string “dd/mm/yyyy” to an Excel date. Excel holds dates as the number of dates since 1/1/1900 CE. Replacing:
.Cells(R, C).Value = Part(N)by .Cells(R, C).Value = CDate(Part(N))ought to do the trick. However, Excel sometimes tries to interpret “dd/mm/yyyy” dates as “mm/dd/yyyy”. I think you will be alright but be aware of this possibility.
Your last step is to convert a date from the CE calendar to the Hijri calendar. This is not just a format issue. The two calendars have different year zeroes and different month lengths. There may be a standard conversion function in your country but there does not appear to be one here in the UK. There is help online so you should be able to find a function that will perform the conversion.
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 /
I have two unordered sets of data here:
blah blah:2020:50::7.1:45
movie blah:blahbah, The:1914:54:
I want to extract all the data to the left of the year (aka, 1915 and 1914).
What excel formula would I use for this?
I tried this formula
=IF(ISNUMBER(SEARCH(":",A1)),MID(A1,SEARCH(":",A1),300),A1)
these were the results below:
: blahblah, The:1914:54::7
:1915:50::7.1:45:
This is because there is a colon in the movie title.
The results I need consistently are:
:1914:54::7.9:17::
:1915:50::7.1:45::
Can someone help with this?
You can use Regular Expressions, make sure you include a reference for it in your VBA editor. The following UDF will do the job.
Function ExtractNumber(cell As Range) As String
ExtractNumber = ""
Dim rex As New RegExp
rex.Pattern = "(:\d{4}:\d{2}::\d\.\d:\d{2}::\d:\d:\d:\d:\d:\d:\d)"
rex.Global = True
Dim mtch As Object, sbmtch As Object
For Each mtch In rex.Execute(cell.Value)
ExtractNumber = ExtractNumber & mtch.SubMatches(0)
Next mtch
End Function
Without VBA:
In reality you don't want to find the : You want to find either :1 or :2 since the year will either start with 1 or 2This formula should do it:
=MID(A1,MIN(IFERROR(FIND(":1",A1,1),9999),IFERROR(FIND(":2",A1),9999)),9999)
Look for a four digit string, in a certain range, bounded by colons.
For example:
=MID(A1,MIN(FIND(":" &ROW(INDIRECT("1900:2100"))&":",A1 &":" &ROW(INDIRECT("1900:2100"))&":")),99)
entered as an array formula by holding down ctrl-shift while hitting Enter would ensure years in the range 1900 to 2100. Change those values as appropriate for your data. The 99 at the end represents the longest possible string. Again, that can be increased as required.
You can use the same approach to return just the left hand part, up to the colon preceding the year:
=LEFT(A1,-1+MIN(FIND(":" &ROW(INDIRECT("1900:2100"))&":",A1 &":" &ROW(INDIRECT("1900:2100"))&":")))
Here is a screen shot, showing the original data in B1:B2, with the results of the first part in B4:B5, and the formula for B4 showing in the formula bar.
The results for the 2nd part are in B7:B9
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 /