PHP Regex IF THEN pattern

Question

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

@AvinashRaj Thanks for pointing that out, my bad. I've added an edit to the origin post. — user1537360
– user1537360, Commented Mar 10, 2015 at 12:54

n-dru · Accepted Answer · 2015-03-10 14:00:31Z

1

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

edited Mar 10, 2015 at 14:00

answered Mar 10, 2015 at 12:18

n-dru

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3 Comments

user1537360 Over a year ago

This works, thanks very much. As I said in my post, I'm new to Regex and still find it confusing. Would you mind explaining briefly what your pattern does that mine doesn't?

n-dru Over a year ago

In first segment I contained that it is zero or one pound sign with at least one digit without trailing "p" OR at least one digit with "p" - this segment is one or zero times. Next is optional point with two digits and optional "p". But wait, I just found a bug. It allows now for 3p.50p. I am updating answer with correction!

user1537360 Over a year ago

Thanks for the fix. I tried to fix it myself but was unsuccessful :)

Robin Curbelo · Accepted Answer · 2015-03-10 12:26:21Z

1

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

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

answered Mar 10, 2015 at 12:26

Robin Curbelo

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Comments

peter_the_oak · Accepted Answer · 2015-03-10 12:28:06Z

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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.

answered Mar 10, 2015 at 12:28

peter_the_oak

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Toto · Accepted Answer · 2015-03-10 12:29:37Z

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How about:

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

answered Mar 10, 2015 at 12:29

Toto

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