regex to extract number and date from string

debugcn 投稿 Dev

Dharmesh Mehta

I'm trying to extract date, percentage or number from string. Strings can be:

the response value 10 (from here I want to extract 10)
the response value 10/12/2014 (from here I want to extract 10/12/2014)
the response value 08/2015 (from here I want to extract 08/2015)

I've written regex as (?:\d{2}\/\d{4}|\d{2}(?:\/\d{2}\/\d{4})?) Regex is satisfying 12/12/2014, 10, 02/2012.

I'm also trying to modifying same regex to get 10, 08/2015 and 10/10/2015 but not getting how to get.

How can this be achieved?

The fourth bird

To match your example data, you could use an alternation matching either 2 digits / 4 digits, or match 2 digits with an optional part that matches 2 digits and 4 digits.

\b(?:\d{2}\/\d{4}|\d{2}(?:\/\d{2}\/\d{4})?)\b

Explanation

\b Word boundary, prevent the word char being part of a larger word
(?: Non capture group
- \d{2}\/\d{4} Match 2 digits/4 digits
- | Or
- \d{2} Match 2 digits
- (?:\/\d{2}\/\d{4})? Optionally match /2 digits/4 digits
) Close group
\b Word boundary

Regex demo

Note that 2 and 4 digits could also match 99 and 9999. If you want to make your match more specific, this page can be helpful https://www.regular-expressions.info/dates.html

const pattern = /\b(?:\d{2}\/\d{4}|\d{2}(?:\/\d{2}\/\d{4})?)\b/;
[
  "the response value 10",
  "the response value 10/12/2014",
  "the response value 08/2015"
].forEach(s => console.log(s.match(pattern)[0]));

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