Excel SEARCH failure on identical strings containing "~" - excel

I have a formula reducing to
=SEARCH("Exchange~server","Exchange~server")
Excel errors on this with "A value used in this formula is of the wrong data type". I know my data can be fixed with
=SEARCH(SUBSTITUTE("Exchange~server","~","-"),SUBSTITUTE("Exchange~server","~","-"))
But, what does excel do to make the first option not work?
Edit: updating title to reflect true issue

Note that:
=FIND("Exchange~server","Exchange~server")
works returning a value of 1
However, when you use SEARCH, wild card characters are allowed in Find_text. The rules for that include one that states that a tilde-x eg (~s) will search for only the s. This is generally used to be able to return the literal wild-card characters * and ?. But it will be true for any character following the tilde.
So your find_text argument reduces to "Exchangeserver" which, of course, cannot be found if you are searching "Exchange~server" with the literal tilde.
To do that, you must change find_text to "Exchange~~server" (double tilde ~~). The ~~ will reduce to the literal ~.
i.e: =SEARCH("Exchange~~server","Exchange~server")

Related

Excel formula that produces one of two options

This is my first StackOverflow question, so apologies if I am unclear.
Currently, my work uses an Excel tracking doc to log project info. The column info is like so:
CELL B1 (Project Number) =IF(B2=""," ",MID(B2,FIND("P2",B2),9))
CELL B2 (Project Name) Client / P2XXXXXXX / Name
Thus, the P2XXXXXXX gets pulled out of B2 and populated into B1.
However, management has recently switched systems, so now, some project numbers have the P2XXXXXXX format and others have a PRJ-XXXXX format.
So we need a formula the produces nothing if the cell is blank and EITHER the P2XXXXXXX number or PRJ-XXXXX number if the cell is not blank.
Is it possible? If any further details are needed, let me know. Thanks in advance!
Well, if the / is always there then this can work:
IF(B2="","",MID(B2,FIND("/",B2,1)+2,9))
assuming the name is always 9 characters.
String Between Two Same Characters
Maybe the next month your company will start using a different first letter or could add more numbers e.g. SPRXXXXXXXXXX. So you could solve this problem by extracting whatever is between those two slashes.
=IF(B2="","",TRIM(MID(B2,FIND("/",B2)+1,FIND("/",B2,FIND("/",B2)+1)-FIND("/",B2)-1)))
Find the first character =FIND("/",B2), but we need the next one:
=FIND("/",B2)+1
Find the second character but search from the postition after the first found:
=FIND("/",B2,FIND("/",B2)+1)
Now get the string between them:
=MID(B2,FIND("/",B2)+1,FIND("/",B2,FIND("/",B2)+1)-FIND("/",B2)-1)
(note how the last minus was 'converted' from a plus to a minus (- + + = -)).
Remove the leading and trailing spaces:
=TRIM(MID(B2,FIND("/",B2)+1,FIND("/",B2,FIND("/",B2)+1)-FIND("/",B2)-1))
Add the condition when the cell is blank:
=IF(B2="","",TRIM(MID(B2,FIND("/",B2)+1,FIND("/",B2,FIND("/",B2)+1)-FIND("/",B2)-1)))
Here's another way using LEFT and RIGHT:
=IF(B2="","",TRIM(LEFT(RIGHT(B2,LEN(B2)-FIND("/",B2)),FIND("/",B2))))
Although you can solve this problem with a combination of slicing, trimming, and complex conditionals, the most expressive and easy to maintain solution is to use regular expressions. Regular expressions have a bit of a learning curve, but there's a great playground website where you can experiment with them, and this page has a pretty good writeup on how regular expressions work in excel.
Specifically, this regular expression addresses the two naming conventions you've highlighted, but it can be updated to support more naming conventions as your company inevitably adds more:
P(RJ-)?((\d){9}|(\d){5})
To break that down from left to right:
P: both patterns start with a "P"
(RJ-)? One pattern follows with "RJ-", but the other doesn't. This is a grouped part of the pattern, and the question mark means that this part of the pattern is optional.
((\d){9}|(\d){5}): by far the nastiest part, but this basically means that there is going to be a sequence of numbers (\d), and there will either be nine of them or five of them. By wrapping the whole thing in parenthesis, they are always the second captured group, no matter the length of the sequence of numbers. This means that you can always extract the project id by looking at the value of the second capture group.
You can also make the expression more generalized by replacing ((\d){9}|(\d){5}) with simply (\d+). That just means "one or more digits." That gives you a much more simplified overall expression of this:
P(RJ-)?(\d+)
Depending on whether or not you care about validating strictly that project ids are 5 OR 9 digits long, that pattern above might be suitable, and it has the benefit of being more flexible. Still, the project ID is in the second captured group.

RegExpFind VBA does not return submatch (capture group), but entire match

In Excel, I have the VBA RegExpFind looking for this pattern:
Fields\("(\w+?)"\)\.Value =
Target:
        Fields("lmlEmployeeID").Value = oRs.Fields("lmpEmployeeID").Value
Formula: =RegExpFind([#Code], [#pat], 1, FALSE, 0 )
Result: Fields("lmlEmployeeID").Value =
Expected result: lmlEmployeeID
Though the pattern involves double-quotes and parentheses, as the pattern is a literal in a cell, I avoided excel character escaping problems - so I thought. Anyway those are outside of the capture group.
As the pattern succeeds, I don't know why the first grouping is not honored.
It may appear that I succeeded in matching the Function name shown, but that is a simple case not using RegExpFind.
I have already looked at these related questions:
RegExp VBA : reluctant quantifier does not return the submatch value
match doesn't return capturing group
Regex doesn't omit quotes when matching text in quotes
VBA: Submatching regex
How to match, but not capture, part of a regex?
EDIT: when applying #toddlson's suggestion, there is a #VALUE! error:
"a value of the formula is of the wrong datatype"
EDIT2: The Add-In I have is different. It's probably out of date:
You can use lookbehind and lookahead, which allow you to specify text before and after your desired match, without including them in the match.
New Pattern - using Lookbehind and Lookahead
(?<=Fields\(")(\w+?)(?="\)\.Value =)
Test string
Fields("lmlEmployeeID").Value = oRs.Fields("lmpEmployeeID").Value
Match
lmlEmployeeID
Try it out!
Important Note:
When setting the pattern in VBA, you'll need to escape the quotation marks otherwise it thinks you're closing and opening a string.
RegExpObject.Pattern = "(?<=Fields\("")(\w+?)(?=""\)\.Value =)"
The problem was really the "wrong" Reg Exp Find function. Many out there have the same name with superficial differences in function name case. I uninstalled the one I had and installed the SEO Tools for Excel (the 15 day trial license). It works (with and without lookback/ahead). I would have thought that Microsoft would have native Reg Exp support for Excel by now.

Excel: Find words of certain length in string?

I have this file where I want to make a conditional check for any cell that contains the letter combination "_SOL", or where the string is followed by any numeric character like "_SOL1524", and stop looking after that. So I don't want matches for "_SOLUTION" or "_SOLothercharactersthannumeric".
So when I use the following formula, I also get results for words like "_SOLUTION":
=IF(ISNUMBER(FIND("_SOL",A1))=TRUE,"Yay","")
How can I avoid this, and only get matches if the match is "_SOL" or "_SOLnumericvalue" (one numeric character)
Clarification: The whole strings may be "Blabla_SOL_BLABLA", "Blabla_SOLUTION_BLABLA" or "Blabla_SOL1524_BLABLA"
Maybe this, which will check if the character after "_SOL" is numeric.
=IF(ISNUMBER(VALUE(MID(A1,FIND("_SOL",A1)+4,1))),"Yay","")
Or, as per OP's request and suggestion, to include the possibility of an underscore after "SOL"
=IF(OR(ISNUMBER(VALUE(MID(A1,FIND("_SOL",A1)+4,1))),ISNUMBER(FIND("_SOL_",A1))),"Yay","")
Here is an alternative way to check if your string contains SOL followed by either nothing or any numeric value up to any characters after SOL:
=IF(COUNT(FILTERXML("<t><s>"&SUBSTITUTE(A1,"_","1</s><s>")&"</s></t>","//s[substring-after(.,'SOL')*0=0]")>0),"Yey","Nay")
Just to use in an unfortunate event where you would encounter SOL1TEXT for example. Or, maybe saver (in case you have text like AEROSOL):
=IF(COUNT(FILTERXML("<t><s>"&SUBSTITUTE(A1,"_","</s><s>")&"</s></t>","//s[translate(.,'1234567890','')='SOL']")>0),"Yey","Nay")
And to prevent that you have text like 123SOL123 you could even do:
=IF(COUNT(FILTERXML("<t><s>"&SUBSTITUTE(A1,"_","1</s><s>")&"</s></t>","//s[starts-with(., 'SOL') and substring(., 4)*0=0]")>0),"Yey","Nay")

Tell vim to add commas to a number, e.g. change 31415926 to 31,415,926

I have a very large number (a couple hundred digits long), and I'd like to use vim to add commas to the number in the appropriate manner, i.e. after each group of three digits, moving from right to left. How can I do this efficiently?
Taken from here
Substitue command that adds commas in the right spot.
:%s/\(\d\)\(\(\d\d\d\)\+\d\#!\)\#=/\1,/g
This uses a zero width lookahead to match any number that isn't followed by groups of three numbers followed by one number. (or 3n+1 numbers)
So the numbers that match in are marked with ^. These are then replaced with a comma after it the match.
31415926
^ ^
Which replaces to
31,415,926
A friend of mine suggests using the printf program: ciw<C-r>=system("printf \"%'d\" ".shellescape(#"))<CR>.
This is one way of doing it:
s/\d\{-1,}\ze\(\d\{3}\)\+\s/&,/g
Notes:
\{-1,} is saying match at least 1 but in a non-greedy way (Vim doesn't seem to support the usual \+\? syntax; also, for quantifiers, you just need to escape the opening curly brace)
\ze is saying match the pattern behind this but don't store the match in & (equivalent to positive look-ahead)
\(\d\{3}\)\+\> matches groups of 3 digits that ends with word-nonword boundary (word in this sense means alphanumerical + underscore).
Alternatively, you can use \s for space/tab, or \D for non-digit instead of \>, whichever fits your needs better
The way that I used is to create a macro that adds one single comma, and then invoke the macro a whole bunch of times, like qahhi,<ESC>hq#a#a#a#a…

How do I get the last character of a string using an Excel function?

How do I get the last character of a string using an Excel function?
No need to apologize for asking a question! Try using the RIGHT function. It returns the last n characters of a string.
=RIGHT(A1, 1)
=RIGHT(A1)
is quite sufficient (where the string is contained in A1).
Similar in nature to LEFT, Excel's RIGHT function extracts a substring from a string starting from the right-most character:
SYNTAX
RIGHT( text, [number_of_characters] )
Parameters or Arguments
text
The string that you wish to extract from.
number_of_characters
Optional. It indicates the number of characters that you wish to extract starting from the right-most character. If this parameter is omitted, only 1 character is returned.
Applies To
Excel 2016, Excel 2013, Excel 2011 for Mac, Excel 2010, Excel 2007, Excel 2003, Excel XP, Excel 2000
Since number_of_characters is optional and defaults to 1 it is not required in this case.
However, there have been many issues with trailing spaces and if this is a risk for the last visible character (in general):
=RIGHT(TRIM(A1))
might be preferred.
Looks like the answer above was a little incomplete try the following:-
=RIGHT(A2,(LEN(A2)-(LEN(A2)-1)))
Obviously, this is for cell A2...
What this does is uses a combination of Right and Len - Len is the length of a string and in this case, we want to remove all but one from that... clearly, if you wanted the last two characters you'd change the -1 to -2 etc etc etc.
After the length has been determined and the portion of that which is required - then the Right command will display the information you need.
This works well combined with an IF statement - I use this to find out if the last character of a string of text is a specific character and remove it if it is. See, the example below for stripping out commas from the end of a text string...
=IF(RIGHT(A2,(LEN(A2)-(LEN(A2)-1)))=",",LEFT(A2,(LEN(A2)-1)),A2)
Just another way to do this:
=MID(A1, LEN(A1), 1)

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