Grab the second number and the word after it using regexp

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I have a string:

str = "1 - hello world 1234 abc 5678 def"

I want to grab 1234 abc from the string.

My current implementation str[/\d+\s*\D*$/] only grabs the last number with the word.

ruby regex
1 Answers

You can use

str[/\d+\D+(\d+\s*\w+)/, 1]
str[/\d+\D+\K\d+\s*\w+/]

See this regex demo and this regex #2 demo.

Here, only the first regex match is searched for. Details:

  • \d+ - one or more digits
  • \D+ - one or more non-digits
  • (\d+\s*\w+) - Group 1: one or more digits, zero or more whitespaces, and one or more word chars
  • \K - match reset operator that discards all text matched so far.

NOTE:

  • If a "word" is a chunk of non-whitespace chars, replace \w+ with \S+
  • If you want to support float values (numbers with fractional parts), you will need to replace \d+ with \d*\.?\d+ or \d+(?:\.\d+)? patterns.

Here is a Ruby test:

str = "1 - hello world 1234 abc 5678 def"
puts str[/\d+\D+(\d+\s*\w+)/, 1]
# => 1234 abc
puts str[/\d+\D+\K\d+\s*\w+/]
# => 1234 abc
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