CODEWITHSUNDEEP
regexruby
avsq1
avsq1

Reputation: 93

Regex / Ruby - split keeping delimiter

I need to split a string containing variables/delimiters, something like;

"Hello %Customer Name% your order number is %Order Number% and will be delivered soon"

Using;

string.split(/%/)
=> ["Hello ", "Customer Name", " your order number is ", "Order Number", " and will be delivered soon"]

Which is close to the requirement, but I'm trying to get to;

["Hello ", "%Customer Name%", " your order number is ", "%Order Number%", " and will be delivered soon"]

So essentially I need to split at % but keep it within the returned fields. I've tried a look ahead/behind with regex but cannot get it quite right.

Upvotes: 3

Views: 156

Answers (2)

Cary Swoveland
Cary Swoveland

Reputation: 110665

Here are three ways that could be done.

str = "%Hello% dear %Cust Name% %your order %Order Nbr% was %lost%"

1. Use String#split

r = /
    (?<=     # begin positive lookbehind
      \A     # match beginning of string
      |      # or
      [ ]    # match a space
    )        # end positive lookbehind
    (?=%)    # positive lookahead asserts next char is '%'
    |        # or
    (?<=%)   # positive lookbehind asserts previous char is '%'
    (?=      # begin a positive lookahead
      [ ]    # match a space
      |      # or
      \z     # match end of string
    )        # end positive lookahead
    /x       # free-spacing regex definition mode

str.split r
  #=> ["%Hello%", " dear ", "%Cust Name%", " ", "%your%", " order ",
  #    "%Order Nbr%", " was ", "%lost%"]

2. Use String#scan

r = /
    %[^%]*%       # match '%', 0+ chars other than '%', '%' 
    |             # or
    (?:           # begin non-capture group#
      (?<=\A)     # positive lookbehind asserts at beginning of string
      |           # or
      (?<=%)      # positive lookbehind asserts previous char is '%'
      (?=[ ])     # positive lookahead asserts next char is a space
    )             # end non-capture group
    [^%]*         # match 0+ chars other than '%' 
    (?=           # begin positive lookahead
      \z          # match end of string
      |           # or
      (?<=[ ])    # assert previous char is a space
      %           # match '%'
    )             # end positive lookahead
    /x            # free-spacing regex definition mode

str.scan r 
  #=> ["%Hello%", " dear ", "%Cust Name%", " ", "%your%", " order ",
  #    "%Order Nbr%", " was ", "%lost%"]

3. Use Enumerable#slice_when

str.each_char.slice_when { |a,b|
  (a == ' ') & (b == '%') || (a == '%') & (b == ' ') }.map(&:join)
  #=> ["%Hello%", " dear ", "%Cust Name%", " ", "%your%", " order ",
  #    "%Order Nbr%", " was ", "%lost%"]

Upvotes: 0

Wiktor Stribiżew
Wiktor Stribiżew

Reputation: 626689

You may use String#split with a pattern like

/(%[^%]*%)/

According to the documentation:

If pattern contains groups, the respective matches will be returned in the array as well.

See the regex demo, it matches and captures into Group 1 a % char, then any 0 or more chars other than %, and then a %.

See a Ruby demo:

s = "Hello %Customer Name% your order number is %Order Number% and will be delivered soon"
p s.split(/(%[^%]*%)/)
# => ["Hello ", "%Customer Name%", " your order number is ", "%Order Number%", " and will be delivered soon"]

Upvotes: 1

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