regex to extract specific rule

user_17171093 0
Viewed 56

I have this string:

"/--shirt-blue-/2346985"

I need to detect '/--' I solve using:

\/-{2}

But I need to detect if the string contains '/--' and the final contains '-/digit'

/--anything-/digit

How can I do that? Thanks.

ruby regex
3 Answers

With your shown samples and attempts please try following regex:

^\/--(?:[^-]*-){2}\/\d+$

Here is the Online demo for above used regex in Ruby.

Explanation: Adding detailed explanation for above shown regex.

^\/--          ##Matching starting /(escaped it) followed by 2 dashes.
(?:[^-]*-){2}  ##In a non-capturing group matching everything till 1st occurrence of - including - and this match 2 times.
               ##So basically this will match till 2 dashes.
\/\d+$         ##Matching literal / followed by 1 or more digits at the end of the value.
user_58665800

You can use

/\/--(.*?)-\/\d+\z/

See the Rubular demo.

Details:

  • \/-- - a /-- string
  • (.*?) - Group 1: any zero or more chars other than line break chars as few as possible
  • -\/ - a -/ string
  • \d+ - one or more digits
  • \z - end of string.

In Ruby,

puts "/--shirt-blue-/2346985"[/\/--(.*?)-\/\d+\z/, 1]

shows shirt-blue.

user_38329700

If there should be at least 1 char other than - in between:

^\/--(?:[^-]+-)+\/\d+\z

Explanation

  • ^ Start of string
  • \/ Match /
  • -- Match literally
  • (?:[^-]+-)+ Repeat 1+ times matching any char except - and then match -
  • \/\d+ Match / and 1+ digits
  • \z End of string

See a regex 101 demo and a Rubular demo

user_54249880
Related