I have a message(m) say "fixes #1 ~needs verification".
I want to identify all numbers that append '#'. So, i scan the text:
issue = m.scan(/[^\#][0-9]+/)
But issue is empty unless the number after # is two digits or > 9 meaning if the message is "fixes #10 ~needs verification" then my issue is 10.
What am i doing wrong here?
You're negating the character class, so your regex is matching (anything that isn't #) followed by one or more digits. 2-digit numbers fit this, but one-digit numbers prefixed by # do not.
Here's what you should do instead:
issue = m.scan(/#[0-9]+/)
Or (credit to this answer):
issue = m.scan(/#\d+/)
"fixes #1 ~needs #12verification" .scan(/#\d+/) #=> ["#1", "#12"]
Take the square brackets off of the # sign:
issue = m.scan(/^#[0-9]+/)
Related
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 :-)
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
I have one path in form of string like this Folder1/File.png
But in this string sometimes if file is hidden or folder is hidden I don't want it to be matched by my regex.
regex = %r{([a-zA-Z0-9_ -]*)\/[^.]+$}
input_path = "Folder_1/.file" # This shouldn't be matched.
input_path = "Folder/file.png" # This should be matched.
But my regex works for first input but its not even matching second one.
You are currently looking for \/[^.]+$, that is a / followed by any character except . until the end. Since the filename+extension format has a . character, it fails to match the second case.
Instead of using [^.]+$, check only that the character following / is not ., and match everything after that:
([a-zA-Z0-9_ -]*)\/[^.].*$
While there are some suggestions here that work, my suggestion would be
\/[^.][^\/\n]+$
It finds a slash, followed by anything but a dot, which in turn is followed by one, or more, of anything but a slash or a newline.
To handle the two lines given as an example,
Folder_1/.file
Folder/file.png
it takes 8 steps.
The suggested ones all work, but ([a-zA-Z0-9_ -]*)\/[^.] takes 75 steps, ([a-zA-Z0-9_ -]*)\/[^.]+\.[^.]+\z 78 steps and ([a-zA-Z0-9_ -]*)\/[^.].*$ takes 77 steps.
This may be totally irrelevant and I may have missed some angle, but I wanted to mention it ;)
Se it here at regex101.
regex = %r{([a-zA-Z0-9_ -]*)\/[^.]}
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/
I want to match strings of the forms:
123
.123
1.123
and I am using the following string for my regex
#"^\\d*(?:\\.\\d+)?$"
However, it matches strings of the following forms as well
1.2.3
1..2..3
123...
What's wrong with my regex? I used the ^ and $ because I don't want the string to contain anything other than the number forms mentioned.
EDIT:
I logged what is matched in the string like 78..7 and found that the match location is 0 and length is 0 with a result of "" being matched. Any ideas? Shouldn't the range location be NSNotFound if the length is 0? I suppose the regex expression is fine then and I can just check for !length but that seems like an unnecessary work around.
Try this regex:
^(?<!\.)\d*(\.\d+)?$
I added a negative look-behind assertion that means that no dot is allowed before that numbers. That should fix your problem.
Description
This regex will find valid positive real numbers with or without a decimal point. like 123, .123, 1.123. The expression can be applied against a string where each value tested is on it's own line or find numbers in the middle of a block of text. It will also allow punctuation like periods and commas directly after the number but won't capture them.
(?<=^|\s)\d*\.?\d+(?=[,.;]?(?:\s|$))
Given Input String:
1.2.3
1..2..3
128...
1234
.123
1.123
1...23
1.2.3
123...
I like kittens 345.23, and version 2.3.4 dogs
Matches are:
1234
.123
1.123
345.23
Does this work for you?
#"^\\d*\\.?\\d+$"
Here it is without escaped backslashes:
^\d*\.?\d+$
My best guess is that rekire is right about the $ symbol not working. If that's the case, then the regex does actually match the empty substring at the start of the string, which explains why it says it's found a match of length 0 at location 0, instead of NSNotFound.
This is REGEX match your strings:
[0-9]*(.){0,1}[0-9]+