正则表达式 / Ruby - 拆分保持分隔符

Question

我需要拆分包含变量/分隔符的字符串，例如；

“您好 %Customer Name% 您的订单号是 %Order Number% 并将很快发货”

使用;

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

这接近要求，但我正在努力达到；

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

所以基本上我需要在 % 处拆分，但将其保留在返回的字段中。 我已经尝试使用正则表达式向前/向后看，但不能完全正确。

Answer 1

您可以将String#split与类似的模式一起使用

/(%[^%]*%)/

根据文档：

如果pattern包含组，则相应的匹配项也将在数组中返回。

请参阅正则表达式演示，它匹配并捕获到第 1 组一个%字符，然后是除%之外的任何 0 个或更多字符，然后是一个% 。

请参阅Ruby 演示：

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"]

Answer 2

以下是可以完成的三种方式。

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

1.使用字符串#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.使用字符串#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. 使用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%"]