using regex to return integer from string within integer range

Question

I am trying to capture the last numbers from "hpercent=" in the string but only if it falls into a range from 1-25.

sample strings:

• ipercent=11 | hpercent=10
• ipercent=21 | hpercent=25
• ipercent=10 | hpercent=42
• ipercent=2 | hpercent=2
• ipercent=15 | hpercent=80
• ipercent=56 | hpercent=89

desired output from "hpercent": 10, 25, 2

I tried:

• Get the last digits: [^=]+(?=$)

• My range for only numbers 1-25: ^[1-9]$|^1[0-9]$|^2[0-5]$

just not sure how to put the above regex together in one expression

UPDATE:

Sorry, very new to this. but I guess I need to capture the last number as a group. but as you can see when it captures the last number it IS beyond 25.

• This captures everything but also the 6 in 26: [1-9]$|^1[0-9]$|^2[0-5]$
• This captures everything but not single digits: ^[1-9]$|1[0-9]$|2[0-5]$

I should only capture between 1-25.

Adobe Classifications

Answer 1

Just to put your regexes together in one expression:

/([1-9]|1[0-9]|2[0-5])(?=$)/

Answer 2

Since you didn't specify what's your regex flavor nor programming language you're using, here's a solution using vim:

/\v\=(\d|1\d|2[0-5])$/

What's going on:

• only capture things comprised between = and EOL ($)
• then capture:
  • a single digit from 1 to 9, or
  • a 1 followed by any digit (so 10 up to 19), or
  • a 2 followed by anything from 0 to 5 (so anything from 20 up to 25)

It should be easy to convert this to your regex flavor.

Answer 3

This should work hpercent=([1-9]|1[0-9]|2[0-5])(?![0-9])

Expanded

 hpercent=
 ( [1-9] | 1 [0-9] | 2 [0-5] )  # (1)
 (?! [0-9] )

Simplified by a number range regex generator tool