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}\/
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 :-)
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 can't figure out what does this regex match:
A: "\\/\\/c\\/(\\d*)"
B: "\\/\\/(\\d*)"
I suppose they are matching some kind of number sequence since \d matches any digit but I'd like to know an example of a string that would be a match for this regex.
The pattern syntax is that specified by ICU. Expressions are created with NSRegularExpression in an iOS app and are correct.
The first matches //c/ + 0 or more digits. The second matches // + 0 or more digits. In both the digits are captured.
An example of a match for A) is //c/123
An example of a match for B) is //12345
When I use Cygwin which emulates Bash on Windows, I sometimes run into situations where I have to escape my escape characters which is what I think is making this expression look so weird. For instance, when I use sed to look for a single '\' I sometimes have to write it as '\\\\'. (Funny, StackOverflow proved my point. If you write 4 backslashes in the comment, it only shows two. So if you process it again, they might all disappear depending on your situation).
Considering this, it might be helpful to think of pairs of backslashes as representing only one if you're coming from a similar situation. My guess would be you are. Because of this I would say Erik Duymelinck is probably spot on. This will capture a sequence of digits that may or may not follow a couple slashes and a c:
//c/000
//00000
This regex matches an odd sequence of characters, which, at first glance, almost seem like a regex, since \d is a digit, and followed by an asterisk (\d*) would mean zero-or-more digits. But it's not a digit, because the escape-slash is escaped.
\\/\\/c\\/(\\d*)
So, for instance, this one matches the following text:
\/\/c\/\
\/\/c\/\d
\/\/c\/\dd
\/\/c\/\ddd
\/\/c\/\dddd
\/\/c\/\ddddd
\/\/c\/\dddddd
...
This one is almost the same
\\/\\/(\\d*)
except you just delete the c\/ from the above results:
\/\/\
\/\/\d
\/\/\dd
\/\/\ddd
\/\/\dddd
\/\/\ddddd
\/\/\dddddd
...
In both cases, the final \ and optional d is [capture group][1] one.
My first impression was that these regexes were intended for escaping in Java strings, meaning they would be completely invalid. If the were escaped for Java strings, such as
Pattern p = Pattern.compile("\\/\\/c\\/(\\d*)");
It would be invalid, because after un-escaping, it would result in this invalid regex:
\/\/c\/(\d*)
The single escape-slashes (\) are invalid. But the \d is valid, as it would mean any digit.
But again, I don't think they're invalid, and they're not escaped for a Java string. They're just odd.
Hi I'm creating a regular expression (ruby) to test the beginning and end of string. I have both parts but can't join them.
Beginning of string
\A(http:\/\/+)
End of string
(.pdf)\z
How to join?
Bonus if it could validate in-between and accept anything (to avoid http://.pdf)
By the way, rubular http://rubular.com is a neat place to validate expressions
Use .+ to match any character except \n one or more times.
\A(http:\/\/+).+(\.pdf)\z
Should match http://www.stackoverflow.com/bestbook.pdf but not http://.pdf
I would like to use regular expression to check if my string have the format like following:
mc_834faisd88979asdfas8897asff8790ds_oa_ids
mc_834fappsd58979asdfas8897asdf879ds_oa_ids
mc_834faispd8fs9asaas4897asdsaf879ds_oa_ids
mc_834faisd8dfa979asdfaspo97asf879ds_dv_ids
mc_834faisd111979asdfas88mp7asf879ds_dv_ids
mc_834fais00979asdfas8897asf87ggg9ds_dv_ids
The format is like mc_<random string>_oa_ids or mc_<random string>_dv_ids . How can I check if my string is in either of these two formats? And please explain the regular expression. thank you.
That's a string start with mc_, while end with _oa_ids or dv_ids, and have some random string in the middle.
P.S. the random string consists of alpha-beta letters and numbers.
What I tried(I have no clue how to check the random string):
/^mc_834faisd88979asdfas8897asff8790ds$_os_ids/
Try this.
^mc_[0-9a-z]+_(dv|oa)_ids$
^ matches at the start of the line the regex pattern is applied to.
[0-9a-z] matces alphabetic and numeric chars.
+ means that there should be one or more chars in this set
(dv|oa) matches dv or oa
$ matches at the end of the string the regex pattern is applied to.
also matches before the very last line break if the string ends with a line break.
Give /\Amc_\w*_(oa|dv)_ids\z/ a try. \A is the beginning of the string, \z the end. \w* are one or more of letters, numbers and underscores and (oa|dv) is either oa or dv.
A nice and simple way to test Ruby Regexps is Rubular, might have a look at it.
This should work
/mc_834([a-z,0-9]*)_(oa|dv)_ids/g
Example: http://regexr.com?2v9q7