CODEWITHSUNDEEP
phpregex
ImLearning
ImLearning

Reputation: 75

REGEX get currency instead of date

I need to read numbers from a string in the format

{any number of digits} {period} {two digits}

Example:

256.23 = 256.23
231.1 = false
2321 = false
das2312.23 = false

However, the problem is that in this string there are dates sometimes and I would like to achieve the following effect:

20.10.20 = false
23.12 = 23.12

I currently have a ReGex like this:

(\d+\.\d{2})

but this return like this:

29.74.23 = 29.74

I also tried:

(\d+\.\d{2}(?!\.))

but it's return = 74.23.

@edit: PHP

Upvotes: 0

Views: 50

Answers (2)

Grismar
Grismar

Reputation: 31354

If the language where you use this supports it, you would use a lookbehind and lookahead, so start with (?<=(?:\s|^)) and end with (?=(?:\s|$)) i.e. before the match there should be the start of the string, or a space, and similar at the end:

(?<=(?:\s|^))(\d+\.\d{2})(?=(?:\s|$))

Of course, if the number would always be the only thing in the string, you could just use:

^(\d+\.\d{2})$

(?<=(?:\s|^)) explained:

  • the (?<=..) is a lookbehind expression, meaning "check that this comes immediately before"
  • the (?:\s|^) just means "match either whitespace or the start of the line" and thanks to the ?: it doesn't become a numbered group.

(?=(?:\s|$)) is similar, except that $ is the end of the line and it starts with ?=, which means "check that this comes immediately after"

@thefourthbird correctly pointed out that you can avoid the non-capturing group in this case (since it's already in the lookbehind/lookahead parentheses):

(?<=\s|^)(\d+\.\d{2})(?=\s|$)

Upvotes: 1

sln
sln

Reputation: 2759

If you don't want surrounding digits or dots and disregarding surrounding dots:

(?<![.\d])\d+\.\d\d(?![\d.])

You can always add to the classes to disregard any other characters not wanted.

But if you go that route, you'd likely need whitespace boundary's instead.

(?<!\S)\d+\.\d\d(?!\S)

which is much more potent.

Upvotes: 0

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