I get an error with my code: =IF(ISNUMBER(SEARCH("$",B3),RIGHT(B3,LEN(B3)-1)*A1000,B3))
The error is: Incorrect number of arguments for IF. Expected between 2 and 3 arguments, but found 1 arguments.
It should convert b3 to euros if there is a "$" before the number.A1000= usd/eur in numbers.
The sheet: https://docs.google.com/spreadsheets/d/1RnYi9H70fqS5wy1OQO2jex7xdNl-1kFLVbm3uDjejTg/edit?usp=sharing .I am trying in G/H 29
try:
=IF(ISNUMBER(SEARCH("$"; B3)); RIGHT(B3; LEN(B3)-1)*A1000; B3)
Related
I'm trying to create a logic as the title but "Wrong number of arguments to IF. Expected between 2 and 3 arguments, but got 1 arguments" what am i doing wrong?
formula:
=IF(OR(U28=1,isnumber(MATCH("TERMINATE",P28,0)),"COMPLETE","PENDING"))
Mismatch of brackets () , try below formula:
=IF(OR(U28=1,isnumber(MATCH("TERMINATE",P28,0))),"COMPLETE","PENDING")
try:
=IF((U28=1)+(ISNUMBER(MATCH("TERMINATE", P28, ))), "COMPLETE", "PENDING")
I have a model which has a column named code, which is a combination of the model's name column and its ID with leading zeros.
name = 'Rocky'
id = 16
I have an after_create callback which runs and generates the code:
update(code: "#{self.name[0..2].upcase}%.4d" % self.id)
The generated code will be:
"ROC0016"
The code is working.
I found (%.4d" % self.id) from another project, but I don't know how it works.
How does it determine the number of zeros to be preceded based on the passed integer.
You’re using a "format specifier". There are many specifiers, but the one you’re using, "%d", is the decimal specifier:
% starts it. 4 means it should always use at least four numbers, so if the number is only two digits, it gets padded with 0s to fill in the rest of the numbers. The second % means replace 4d with whatever comes after it. So in your case, 4d is getting replaced with "0016".
sprintf has more information about format specifiers.
You can read more about String#% in the documentation also.
After the percentage sign ("%") is a decimal (".") and a number. That number is the number of total digits in the result. If the result is less than this value, additional zeros will be added.
Thus, in this first example, the result is "34" but length was set to "4". The result will have two leading zeros to fill it into four digits.
"This is test string %.4d" % 34
result => "This is test string 0034"
"I want more zeroes in my code %.7d" % 34
result => "I want more zeroes in my code 0000034"
When trying to get an output of "#.## M" I receive an error, but when I remove the M, I do not. Having the M at the end is imperative to the end result. Formatting alone outside of the formula does not work.
I've tried several web searches that did not fix the issue. I don't know what else to attempt.
=(TEXT('National Economy & Finance'!C13, "#.## M"))
I expect an output of 42.42 M but instead receive the following error:
Invalid format pattern '#.## M' in TEXT evaluation.
You just need to escape the M:
=(TEXT('National Economy & Finance'!C13, "#.## \M"))
"#.##" would be interpreted as a decimal number and "M" would normally be interpreted as month, so you can't have both together in the same format.
You can also just put
=(TEXT('National Economy & Finance'!C13, "#.##"))&" M"
Trying to convert below Phone numbers patterns into single pattern with open and close brackets for first 3 digits using formula only in Google spreadsheet.
214-7671378
214660-9212
214.412-2034
(972) 223-6473
214 502 7196
To
(214) 767-1378
(214) 660-9212
(214) 412-2034
(972) 223-6473
(214) 502-7196
In google-spreadsheet
=text(REGEXREPLACE(TEXT(A1,"#"),"\D+",""),"(###) ###-####")
If i have a string like "123123123" - Here 123 is repeated 3 times.
1. So how can i get only "123" in ruby?
2. So if the string is "12312312" - Here 123 is repeated 2 times and then just 12, so here still i need to get "123".
3. Even if string is 99123123123, still i need to get 123.
Is this possible in Ruby Regex?
EDIT: I want this to solve Project Euler Problem 26 . So here 123 can be anything. All i want is to extract 1 number of at-least 2 repeated numbers.
This regex will detect all repeating groups.
(\d+)(?=.*\1)
Demo
Works great with ruby too.
result = '9912341234123'.scan(/(\d+)(?=.*\1)/)
#gets group with largest length
longestRepeatingGroup = result.max_by{|arr| arr[0].length}
puts longestRepeatingGroup
puts longestRepeatingGroup[0].length
Try this
99123123123.scan(/123/).count
12312312.scan(/123/).count