Grab the second number and the word after it using regexp

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.

>Solution :

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