I have varibale which contains raw data as shown below.
I want to replace the comma inside double quotes to nothing/blank.
I used replaceAll(',',''), but all other commas are also getting replaced.
So need regex function to identify pattern like "123,456,775" and then replace here comma into blank.
var = '
2/5/2023,25,"717,990","18,132,406"
2/4/2023,27,"725,674","19,403,116"
2/3/2023,35,"728,501","25,578,008"
1/31/2023,37,"716,580","26,358,186"
2/1/2023,37,"720,466","26,494,010"
1/30/2023,37,"715,685","26,517,878"
2/2/2023,37,"723,545","26,603,765" '
Tried replaceAll, but did not work
If you just want to replace "," with "", you have to escape the quotes this will do:
var.replaceAll(/\",\"/, /\"\"/)
If you want to replace commas inside the number strings, "725,674" with "725674" you will have to use a regex and capture groups, like this:
var.replaceAll(/(\"\d+),(\d+\")/, /$1$2/)
It will change for three groupings, like "18,132,406", you will have to use three capture groups.
I had to split string data based on Comma.
This is the excel data:-
Please find the excel data
string strCurrentLine="\"Himalayan Salt Body Scrub with Lychee Essential Oil from Majestic Pure, All Natural Scrub to Exfoliate & Moisturize Skin, 12 oz\",SKU_27,\"Tombow Dual Brush Pen Art Markers, Portrait, 6-Pack\",SKU_27,My Shopify Store 1,Valid,NonInventory".
Regex CSVParser = new Regex(",(?=(?:[^\"]\"[^\"]\")(?![^\"]\"))");
string[] lstColumnValues = CSVParser.Split(strCurrentLine);
I have attached the image.The problem is I used the Regex to split the string with comma but i need the ouptut just like SKU_27 because string[0] and string2 contains the forward and backward slash.I need the output string1 and remove the forward and backward slash.
The file seems to be a CVA file. For CVA to be properly formatted, it will use quotes "" to wrap strings that contains comma, such as
id, name, date
1,"Some text, that includes comma", 2020/01/01
Simply split the string by comma, you will get the 2nd column with double quote.
I'm not sure whether you are asking how to remove the double-quotes from lstColumnValues[0] and lstColumnValues[2], or add them to lstColumnValues[1].
To remove the double-quotes, just use Replace:
string myString = lstColumnValues[0].Replace("\"", "");
If you need to add them:
string myString = $"\"{lstColumnValues[1]}\"";
I try to import a CSV file in Excel, using ; as delimiters, but some columns contains
; and/or quotes.
My problem is : I can use double quotes to ignore the delimiters for a specific string, but if there is a double quote inside the string, it ignores delimiters until the first double quote, but not after.
I don't know if it's clear, it's not that easy to explain.
I will try to explain with a example :
Suppose I have this string this is a;test : I use double quotes around the string, to ignore the delimiter => It works.
Now if this string contains delimiters AND double quotes : my trick doesn't work anymore. For example if I have the string this; is" a;test : My added double quotes around the string ignore delimiters for the first part (the delimiter in the part this; is is correctly ignored, but since there is a double quote after, Excel doesn't ignore the next delimiter in the a;test part.
I tried my best to be as clear as possible, I hope you'll understand what is the problem.
When reading in a quoted string in a csv file, Excel will interpret all pairs of double-quotes ("") with single double-quotes(").
so "this; is"" a;test" will be converted to one cell containing this; is" a;test
So replace all double-quotes in your strings with pairs of double quotes.
Excel will reverse this process when exporting as CSV.
Here is some CSV
a,b,c,d,e
"""test1""",""",te"st2,"test,3",test"4,test5
And this is how it looks after importing into Excel:
Import your Excel file in openOffice and export as CSV (column escaped with " unlike Excel csv, utf8, comma against ";").
I have an Excel spreadsheet containing a list of strings. Each string is made up of several words, but the number of words in each string is different.
Using built in Excel functions (no VBA), is there a way to isolate the last word in each string?
Examples:
Are you classified as human? -> human?
Negative, I am a meat popsicle -> popsicle
Aziz! Light! -> Light!
This one is tested and does work (based on Brad's original post):
=RIGHT(A1,LEN(A1)-FIND("|",SUBSTITUTE(A1," ","|",
LEN(A1)-LEN(SUBSTITUTE(A1," ","")))))
If your original strings could contain a pipe "|" character, then replace both in the above with some other character that won't appear in your source. (I suspect Brad's original was broken because an unprintable character was removed in the translation).
Bonus: How it works (from right to left):
LEN(A1)-LEN(SUBSTITUTE(A1," ","")) – Count of spaces in the original string
SUBSTITUTE(A1," ","|", ... ) – Replaces just the final space with a |
FIND("|", ... ) – Finds the absolute position of that replaced | (that was the final space)
Right(A1,LEN(A1) - ... )) – Returns all characters after that |
EDIT: to account for the case where the source text contains no spaces, add the following to the beginning of the formula:
=IF(ISERROR(FIND(" ",A1)),A1, ... )
making the entire formula now:
=IF(ISERROR(FIND(" ",A1)),A1, RIGHT(A1,LEN(A1) - FIND("|",
SUBSTITUTE(A1," ","|",LEN(A1)-LEN(SUBSTITUTE(A1," ",""))))))
Or you can use the =IF(COUNTIF(A1,"* *") syntax of the other version.
When the original string might contain a space at the last position add a trim function while counting all the spaces: Making the function the following:
=IF(ISERROR(FIND(" ",B2)),B2, RIGHT(B2,LEN(B2) - FIND("|",
SUBSTITUTE(B2," ","|",LEN(TRIM(B2))-LEN(SUBSTITUTE(B2," ",""))))))
This is the technique I've used with great success:
=TRIM(RIGHT(SUBSTITUTE(A1, " ", REPT(" ", 100)), 100))
To get the first word in a string, just change from RIGHT to LEFT
=TRIM(LEFT(SUBSTITUTE(A1, " ", REPT(" ", 100)), 100))
Also, replace A1 by the cell holding the text.
A more robust version of Jerry's answer:
=TRIM(RIGHT(SUBSTITUTE(TRIM(A1), " ", REPT(" ", LEN(TRIM(A1)))), LEN(TRIM(A1))))
That works regardless of the length of the string, leading or trailing spaces, or whatever else and it's still pretty short and simple.
I found this on google, tested in Excel 2003 & it works for me:
=IF(COUNTIF(A1,"* *"),RIGHT(A1,LEN(A1)-LOOKUP(LEN(A1),FIND(" ",A1,ROW(INDEX($A:$A,1,1):INDEX($A:$A,LEN(A1),1))))),A1)
[edit] I don't have enough rep to comment, so this seems the best place...BradC's answer also doesn't work with trailing spaces or empty cells...
[2nd edit] actually, it doesn't work for single words either...
=RIGHT(TRIM(A1),LEN(TRIM(A1))-FIND(CHAR(7),SUBSTITUTE(" "&TRIM(A1)," ",CHAR(7),
LEN(TRIM(A1))-LEN(SUBSTITUTE(" "&TRIM(A1)," ",""))+1))+1)
This is very robust--it works for sentences with no spaces, leading/trailing spaces, multiple spaces, multiple leading/trailing spaces... and I used char(7) for the delimiter rather than the vertical bar "|" just in case that is a desired text item.
This is very clean and compact, and works well.
{=RIGHT(A1,LEN(A1)-MAX(IF(MID(A1,ROW(1:999),1)=" ",ROW(1:999),0)))}
It does not error trap for no spaces or one word, but that's easy to add.
Edit:
This handles trailing spaces, single word, and empty cell scenarios. I have not found a way to break it.
{=RIGHT(TRIM(A1),LEN(TRIM(A1))-MAX(IF(MID(TRIM(A1),ROW($1:$999),1)=" ",ROW($1:$999),0)))}
=RIGHT(A1,LEN(A1)-FIND("`*`",SUBSTITUTE(A1," ","`*`",LEN(A1)-LEN(SUBSTITUTE(A1," ","")))))
New answer 9/28/2022
Considering the new excel function: TEXTAFTER (check availability) you can achieve it with a simple formula:
=TEXTAFTER(A1," ", -1)
To add to Jerry and Joe's answers, if you're wanting to find the text BEFORE the last word you can use:
=TRIM(LEFT(SUBSTITUTE(TRIM(A1), " ", REPT(" ", LEN(TRIM(A1)))), LEN(SUBSTITUTE(TRIM(A1), " ", REPT(" ", LEN(TRIM(A1)))))-LEN(TRIM(A1))))
With 'My little cat' in A1 would result in 'My little' (where Joe and Jerry's would give 'cat'
In the same way that Jerry and Joe isolate the last word, this then just gets everything to the left of that (then trims it back)
Copy into a column, select that column and HOME > Editing > Find & Select, Replace:
Replace All.
There is a space after the asterisk.
Imagine the string could be reversed. Then it is really easy. Instead of working on the string:
"My little cat" (1)
you work with
"tac elttil yM" (2)
With =LEFT(A1;FIND(" ";A1)-1) in A2 you get "My" with (1) and "tac" with (2), which is reversed "cat", the last word in (1).
There are a few VBAs around to reverse a string. I prefer the public VBA function ReverseString.
Install the above as described. Then with your string in A1, e.g., "My little cat" and this function in A2:
=ReverseString(LEFT(ReverseString(A1);IF(ISERROR(FIND(" ";A1));
LEN(A1);(FIND(" ";ReverseString(A1))-1))))
you'll see "cat" in A2.
The method above assumes that words are separated by blanks. The IF clause is for cells containing single words = no blanks in cell. Note: TRIM and CLEAN the original string are useful as well. In principle it reverses the whole string from A1 and simply finds the first blank in the reversed string which is next to the last (reversed) word (i.e., "tac "). LEFT picks this word and another string reversal reconstitutes the original order of the word (" cat"). The -1 at the end of the FIND statement removes the blank.
The idea is that it is easy to extract the first(!) word in a string with LEFT and FINDing the first blank. However, for the last(!) word the RIGHT function is the wrong choice when you try to do that because unfortunately FIND does not have a flag for the direction you want to analyse your string.
Therefore the whole string is simply reversed. LEFT and FIND work as normal but the extracted string is reversed. But his is no big deal once you know how to reverse a string. The first ReverseString statement in the formula does this job.
=LEFT(A1,FIND(IF(
ISERROR(
FIND("_",A1)
),A1,RIGHT(A1,
LEN(A1)-FIND("~",
SUBSTITUTE(A1,"_","~",
LEN(A1)-LEN(SUBSTITUTE(A1,"_",""))
)
)
)
),A1,1)-2)
I translated to PT-BR, as I needed this as well.
(Please note that I've changed the space to \ because I needed the filename only of path strings.)
=SE(ÉERRO(PROCURAR("\",A1)),A1,DIREITA(A1,NÚM.CARACT(A1)-PROCURAR("|", SUBSTITUIR(A1,"\","|",NÚM.CARACT(A1)-NÚM.CARACT(SUBSTITUIR(A1,"\",""))))))
Another way to achieve this is as below
=IF(ISERROR(TRIM(MID(TRIM(D14),SEARCH("|",SUBSTITUTE(TRIM(D14)," ","|",LEN(TRIM(D14))-LEN(SUBSTITUTE(TRIM(D14)," ","")))),LEN(TRIM(D14))))),TRIM(D14),TRIM(MID(TRIM(D14),SEARCH("|",SUBSTITUTE(TRIM(D14)," ","|",LEN(TRIM(D14))-LEN(SUBSTITUTE(TRIM(D14)," ","")))),LEN(TRIM(D14)))))
You can achieve this also by reversing the string and finding the first space
=MID(C3,2+LEN(C3)-SEARCH(" ",CONCAT(MID(C3,SEQUENCE(LEN(C3),,LEN(C3),-1),1))),LEN(A1))
Reverse the string
CONCAT(MID(C3,SEQUENCE(LEN(C3),,LEN(C3),-1),1))
Find the first space in the reversed string
SEARCH(" ",...
Take the position of the space found in the reversed string off the length of the string and return that portion
=MID(C3,2+LEN(C3)-SEARCH...
I also had a task like this and when I was done, using the above method, a new method occured to me: Why don't you do this:
Reverse the string ("string one" becomes "eno gnirts").
Use the good old Find (which is hardcoded for left-to-right).
Reverse it into readable string again.
How does this sound?