Split string by whitespaces, ignoring escaped whitespaces [duplicate]

Question

I've been using the Percentage Strings Literal to convert strings like "one two three four\ five" into an array.

%w(one two three four\ five)

returns:

["one", "two", "three", "four five"]

Now I want to do this dynamically, so can use Literals anymore.

Which regex pattern can I use to convert my string above into an array?

I'm looking for a regex pattern to place into a ruby split method that will take "one two three four\ five" and return ["one", "two", "three", "four five"].

Note: I only want to split by whitespaces that aren't escaped, like above. Four and Five were combined into the same string because the whitespace that separated them was escaped.

Answer 1

If your strings have no escape sequences, you may use a splitting approach with

.split(/(?<!\\)\s+/)

Here, (?<!\\)\s+ matches 1+ whitespaces (\s+) that are not preceded with \.

If your strings may contain escape sequences, a matching approach is preferable as it is more reliable:

.scan(/(?:[^\\\s]|\\.)+/)

See the Ruby demo.

It will match 1 or more characters other than \ and whitespace (with [^\\\s]) and any escape sequence (matched with \\., a backslash + any char other than line break chars).

To get rid of \ symbols, you will have to use a gsub later.