Getting a substring in Ruby by x number of chars
RubyRuby Problem Overview
I'm trying to produce some Ruby code that will take a string and return a new one, with a number x number of characters removed from its end - these can be actual letters, numbers, spaces etc.
Ex: given the following string
a_string = "a1wer4zx"
I need a simple way to get the same string, minus - say - the 3 last characters. In the case above, that would be "a1wer". The way I'm doing it right now seems very convoluted:
an_array = a_string.split(//,(a_string.length-2))
an_array.pop
new_string = an_array.join
Any ideas?
Ruby Solutions
Solution 1 - Ruby
How about this?
s[0, s.length - 3]
Or this
s[0..-4]
edit
s = "abcdefghi"
puts s[0, s.length - 3] # => abcdef
puts s[0..-4] # => abcdef
Solution 2 - Ruby
Use something like this:
s = "abcdef"
new_s = s[0..-2] # new_s = "abcde"
See the slice
method here: <http://ruby-doc.org/core/classes/String.html>
Solution 3 - Ruby
Another option could be to use the slice
method
a_string = "a1wer4zx"
a_string.slice(0..5)
=> "a1wer4"
Documentation: http://ruby-doc.org/core-2.5.0/String.html#method-i-slice
Solution 4 - Ruby
Another option is getting the list of chars
of the string, take
ing x chars and join
ing back to a string:
[13] pry(main)> 'abcdef'.chars.take(2).join
=> "ab"
[14] pry(main)> 'abcdef'.chars.take(20).join
=> "abcdef"
Solution 5 - Ruby
if you need it in rails you can use first (source code)
s = '1234567890'
x = 4
s.first(s.length - x) # => "123456"
there is also last (source code)
s.last(2) # => "90"
alternatively check from/to