Rails strip all except numbers commas and decimal points - ruby-on-rails

Hi I've been struggling with this for the last hour and am no closer. How exactly do I strip everything except numbers, commas and decimal points from a rails string? The closest I have so far is:-
rate = rate.gsub!(/[^0-9]/i, '')
This strips everything but the numbers. When I try add commas to the expression, everything is getting stripped. I got the aboves from somewhere else and as far as I can gather:
^ = not
Everything to the left of the comma gets replaced by what's in the '' on the right
No idea what the /i does
I'm very new to gsub. Does anyone know of a good tutorial on building expressions?
Thanks

Try:
rate = rate.gsub(/[^0-9,\.]/, '')
Basically, you know the ^ means not when inside the character class brackets [] which you are using, and then you can just add the comma to the list. The decimal needs to be escaped with a backslash because in regular expressions they are a special character that means "match anything".
Also, be aware of whether you are using gsub or gsub!
gsub! has the bang, so it edits the instance of the string you're passing in, rather than returning another one.
So if using gsub! it would be:
rate.gsub!(/[^0-9,\.]/, '')
And rate would be altered.
If you do not want to alter the original variable, then you can use the version without the bang (and assign it to a different var):
cleaned_rate = rate.gsub!(/[^0-9,\.]/, '')
I'd just google for tutorials. I haven't used one. Regexes are a LOT of time and trial and error (and table-flipping).
This is a cool tool to use with a mini cheat-sheet on it for ruby that allows you to quickly edit and test your expression:
http://rubular.com/

You can just add the comma and period in the square-bracketed expression:
rate.gsub(/[^0-9,.]/, '')
You don't need the i for case-insensitivity for numbers and symbols.
There's lots of info on regular expressions, regex, etc. Maybe search for those instead of gsub.

You can use this:
rate = rate.gsub!(/[^0-9\.\,]/g,'')
Also check this out to learn more about regular expressions:
http://www.regexr.com/

Related

How to specify a range in Ruby

I've been looking for a good way to see if a string of items are all numbers, and thought there might be a way of specifying a range from 0 to 9 and seeing if they're included in the string, but all that I've looked up online has really confused me.
def validate_pin(pin)
(pin.length == 4 || pin.length == 6) && pin.count("0-9") == pin.length
end
The code above is someone else's work and I've been trying to identify how it works. It's a pin checker - takes in a set of characters and ensures the string is either 4 or 6 digits and all numbers - but how does the range work?
When I did this problem I tried to use to_a? Integer and a bunch of other things including ranges such as (0..9) and ("0..9) and ("0".."9") to validate a character is an integer. When I saw ("0-9) it confused the heck out of me, and half an hour of googling and youtube has only left me with regex tutorials (which I'm interested in, but currently just trying to get the basics down)
So to sum this up, my goal is to understand a more semantic/concise way to identify if a character is an integer. Whatever is the simplest way. All and any feedback is welcome. I am a new rubyist and trying to get down my fundamentals. Thank You.
Regex really is the right way to do this. It's specifically for testing patterns in strings. This is how you'd test "do all characters in this string fall in the range of characters 0-9?":
pin.match(/\A[0-9]+\z/)
This regex says "Does this string start and end with at least one of the characters 0-9, with nothing else in between?" - the \A and \z are start-of-string and end-of-string matchers, and the [0-9]+ matches any one or more of any character in that range.
You could even do your entire check in one line of regex:
pin.match(/\A([0-9]{4}|[0-9]{6})\z/)
Which says "Does this string consist of the characters 0-9 repeated exactly 4 times, or the characters 0-9, repeated exactly 6 times?"
Ruby's String#count method does something similar to this, though it just counts the number of occurrences of the characters passed, and it uses something similar to regex ranges to allow you to specify character ranges.
The sequence c1-c2 means all characters between c1 and c2.
Thus, it expands the parameter "0-9" into the list of characters "0123456789", and then it tests how many of the characters in the string match that list of characters.
This will work to verify that a certain number of numbers exist in the string, and the length checks let you implicitly test that no other characters exist in the string. However, regexes let you assert that directly, by ensuring that the whole string matches a given pattern, including length constraints.
Count everything non-digit in pin and check if this count is zero:
pin.count("^0-9").zero?
Since you seem to be looking for answers outside regex and since Chris already spelled out how the count method was being implemented in the example above, I'll try to add one more idea for testing whether a string is an Integer or not:
pin.to_i.to_s == pin
What we're doing is converting the string to an integer, converting that result back to a string, and then testing to see if anything changed during the process. If the result is =>true, then you know nothing changed during the conversion to an integer and therefore the string is only an Integer.
EDIT:
The example above only works if the entire string is an Integer and won’t properly deal with leading zeros. If you want to check to make sure each and every character is an Integer then do something like this instead:
pin.prepend(“1”).to_i.to_s(1..-1) == pin
Part of the question seems to be exactly HOW the following portion of code is doing its job:
pin.count("0-9")
This piece of the code is simply returning a count of how many instances of the numbers 0 through 9 exist in the string. That's only one piece of the relevant section of code though. You need to look at the rest of the line to make sense of it:
pin.count("0-9") == pin.length
The first part counts how many instances then the second part compares that to the length of the string. If they are equal (==) then that means every character in the string is an Integer.
Sometimes negation can be used to advantage:
!pin.match?(/\D/) && [4,6].include?(pin.length)
pin.match?(/\D/) returns true if the string contains a character other than a digit (matching /\D/), in which case it it would be negated to false.
One advantage of using negation here is that if the string contains a character other than a digit pin.match?(/\D/) would return true as soon as a non-digit is found, as opposed to methods that examine all the characters in the string.

How to remove from string before __

I am building a Rails 5.2 app.
In this app I got outputs from different suppliers (I am building a webshop).
The name of the shipping provider is in this format:
dhl_freight__233433
It could also be in this format:
postal__US-320202
How can I remove all that is before (and including) the __ so all that remains are the things after the ___ like for example 233433.
Perhaps some sort of RegEx.
A very simple approach would be to use String#split and then pick the second part that is the last part in this example:
"dhl_freight__233433".split('__').last
#=> "233433"
"postal__US-320202".split('__').last
#=> "US-320202"
You can use a very simple Regexp and a ask the resulting MatchData for the post_match part:
p "dhl_freight__233433".match(/__/).post_match
# another (magic) way to acces the post_match part:
p $'
Postscript: Learnt something from this question myself: you don't even have to use a RegExp for this to work. Just "asddfg__qwer".match("__").post_match does the trick (it does the conversion to regexp for you)
r = /[^_]+\z/
"dhl_freight__233433"[r] #=> "233433"
"postal__US-320202"[r] #=> "US-320202"
The regular expression matches one or more characters other than an underscore, followed by the end of the string (\z). The ^ at the beginning of the character class reads, "other than any of the characters that follow".
See String#[].
This assumes that the last underscore is preceded by an underscore. If the last underscore is not preceded by an underscore, in which case there should be no match, add a positive lookbehind:
r = /(?<=__[^_]+\z/
This requires the match to be preceded by two underscores.
There are many ruby ways to extract numbers from string. I hope you're trying to fetch numbers out of a string. Here are some of the ways to do so.
Ref- http://www.ruby-forum.com/topic/125709
line.delete("^0-9")
line.scan(/\d/).join('')
line.tr("^0-9", '')
In the above delete is the fastest to trim numbers out of strings.
All of above extracts numbers from string and joins them. If a string is like this "String-with-67829___numbers-09764" outut would be like this "6782909764"
In case if you want the numbers split like this ["67829", "09764"]
line.split(/[^\d]/).reject { |c| c.empty? }
Hope these answers help you! Happy coding :-)

Lua pattern help (Double parentheses)

I have been coding a program in Lua that automatically formats IRC logs from a roleplay. In the roleplay logs there is a specific guideline for "Out of character" conversation, which we use double parentheses for. For example: ((<Things unrelated to roleplay go here>)). I have been trying to have my program remove text between double brackets (and including both brackets). The code is:
ofile = io.open("Output.txt", "w")
rfile = io.open("Input.txt", "r")
p = rfile:read("*all")
w = string.gsub(p, "%(%(.*?%)%)", "")
ofile:write(w)
The pattern here is > "%(%(.*?%)%)" I've tried multiple variations of the pattern. All resulted in fruitless results:
1. %(%(.*?%)%) --Wouldn't do anything.
2. %(%(.*%)%) --Would remove *everything* after the first OOC message.
Then, my friend told me that prepending the brackets with percentages wouldn't work, and that I had to use backslashes to 'escape' the parentheses.
3. \(\(.*\)\) --resulted in the output file being completely empty.
4. (\(\(.*\)\)) --Same result as above.
5. (\(\(.*?\)\) --would for some reason, remove large parts of the text for no apparent reason.
6. \(\(.*?\)\) --would just remove all the text except for the last line.
The short, absolute question:
What pattern would I need to use to remove all text between double parentheses, and remove the double parentheses themselves too?
You're friend is thinking of regular expressions. Lua patterns are similar, but different. % is the correct escape character.
Your pattern should be %(%(.-%)%). The - is similar to * in that it matches any number of the preceding sequence, but while * tries to match as many characters as it can (it's greedy), - matches the least amount of characters possible (it's non-greedy). It won't go overboard and match extra double-close-parenthesis.

Regex in Ruby not working

I have a string from which I want to extract a certain part:
Original String: /abc/d7_t/g-12/jkl/m-n3/pqr/stu/vwx
Result Desired: /abc/d7_t/g-12/jkl/
The number of characters can vary in the entire string. It has alphabets, numbers, underscore and hyphen. I want to basically cut the string after the 5th "/"
I tried a few regex, but it seems there is some mistake with the format.
If a non-regexp approach is acceptable, how about this:
s.split('/').take(n).join('/')+'/'
Where s if your string (in your case: /abc/d7_t/g-12/jkl/m-n3/pqr/stu/vwx).
def cut_after(s, n)
s.split('/').take(n).join('/')+'/'
end
Then
cut_after("/abc/d7_t/g-12/jkl/m-n3/pqr/stu/vwx", 5)
should work. Not as compact as a regexp, but some people may find it clearer.
The regexp would be: %r(/(?:[^/]+/){4}). Note that it is a good idea in this case to use the %r literal version to avoid escaping slashes. Unescaped slashes are likely the cause of your format errors.
Match any sequence of chars except '/' 4 times :-
(\/[^\/]+){4}\/

Regular expression in Ruby

Could anybody help me make a proper regular expression from a bunch of text in Ruby. I tried a lot but I don't know how to handle variable length titles.
The string will be of format <sometext>title:"<actual_title>"<sometext>. I want to extract actual_title from this string.
I tried /title:"."/ but it doesnt find any matches as it expects a closing quotation after one variable from opening quotation. I couldn't figure how to make it check for variable length of string. Any help is appreciated. Thanks.
. matches any single character. Putting + after a character will match one or more of those characters. So .+ will match one or more characters of any sort. Also, you should put a question mark after it so that it matches the first closing-quotation mark it comes across. So:
/title:"(.+?)"/
The parentheses are necessary if you want to extract the title text that it matched out of there.
/title:"([^"]*)"/
The parentheses create a capturing group. Inside is first a character class. The ^ means it's negated, so it matches any character that's not a ". The * means 0 or more. You can change it to one or more by using + instead of *.
I like /title:"(.+?)"/ because of it's use of lazy matching to stop the .+ consuming all text until the last " on the line is found.
It won't work if the string wraps lines or includes escaped quotes.
In programming languages where you want to be able to include the string deliminator inside a string you usually provide an 'escape' character or sequence.
If your escape character was \ then you could write something like this...
/title:"((?:\\"|[^"])+)"/
This is a railroad diagram. Railroad diagrams show you what order things are parsed... imagine you are a train starting at the left. You consume title:" then \" if you can.. if you can't then you consume not a ". The > means this path is preferred... so you try to loop... if you can't you have to consume a '"' to finish.
I made this with https://regexper.com/#%2Ftitle%3A%22((%3F%3A%5C%5C%22%7C%5B%5E%22%5D)%2B)%22%2F
but there is now a plugin for Atom text editor too that does this.

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