How to extract percentage text using regex?

Question

From a text file I have to find all the line which have containing percentages (numbers + % or percent) for this I have created a regex but it doesnot work correctly.

regex string: \b(\d+(%|(percent)))\b

and my inputs are

• 423%
• 423%:
• 10percent
• 10percent:

for 1st two cases it is not matching but for 3rd and 4th it works.

My requirement is to identfy the line which having numbers + % or percent and around it there should be no alphabate or number

Answer 1

The word boundary after % prevents it from matching before non-word chars.

Use

\b\d+(?:%|percent\b)

See the regex demo

The pattern matches:

• \b - a leading word boundary
• \d+ - 1+ digits
• (?:%|percent\b) - one of the two alternatives:
  • % - a percentage sign
  • percent\b - word percent followed with a word boundary.