CODEWITHSUNDEEP
phpregex
user1537360
user1537360

Reputation: 4811

PHP Regex IF THEN pattern

I'm new to writing Regex patterns and I'm struggling to understand why the following line doesn't work.

/^(£)?[0-9]+(?(?=\.[0-9]{2}){0,1}(p)?|$)/

Note: I'm writing this in PHP

I want the code to find £3.10p, but not £3p. Essentially, the letter 'p' can't be allowed unless it is preceded with a decimal point and 2 digits.

EDIT: To clarify, the letter p can be used at the end of the string, however if the string contains a £ and/or a decimal point, the p must be preceded by the point and 2 digits.

More examples of valid inputs: £3.50 350 £350 234p

Invalid input: £2p

Could someone please fix this and explain where I've gone wrong here?

Thanks

Upvotes: 1

Views: 99

Answers (4)

n-dru
n-dru

Reputation: 9430

If 0.50p is allowed, then you can do it like this:

^((£?[0-9]+)(?!p)|([0-9]+p?))?(?<!p)(\.[0-9]{2})?p?$

Regex saved with all your examples here: https://regex101.com/r/rE1bT9/3

Upvotes: 1

Toto
Toto

Reputation: 91498

How about:

/^(?:£?[0-9]+(?:\.[0-9]{2})?|[0-9]+p?)$)/

Upvotes: 0

peter_the_oak
peter_the_oak

Reputation: 3710

It is unclear how your valid sample "234p" matches your rule "p is allowed if there are at least two digits and a point". However, in your question you are using positive lookahead, this seems an overhead here.

Your rule for p may be written as: (\.[0-9]{2}p?)

So over all, you just need: /^(£)?[0-9]+(\.[0-9]{2}p?)$/

And if you allow "234p" also, just make the period optional: /^(£)?[0-9]+(\.?[0-9]{2}p?)$/

Try it out here: http://www.regexr.com/

The latter regex gives positive feedback to all your valid samples, and it denies the invalid input. It is unclear what should happen if there are only two digits, and if it is important to catch some pieces, there should be more brackets.

Upvotes: 0

Robin Curbelo
Robin Curbelo

Reputation: 1343

Try this: /^(?(?=£)(£\d+\.\d{2}p?|£\d+)|\d+p?)$/

You can test it here: https://regex101.com/r/mG8kR0/1

Upvotes: 1

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