CODEWITHSUNDEEP
javascriptregex
CJLopez
CJLopez

Reputation: 5815

Regex to get complete optional part if present

I need to verify if a entered order number either is an original or a duplicate.

Order number can have either numbers or character and must be 13 ([A-Z0-9]{13})

A duplicate, its sepatated by a dash and its added a single capital letter ([-]{1}[A-Z]{1})

Given that the user can either enter the original or a duplicate order, the duplicate can be present or not, so, so far, I have this

([a-zA-Z0-9]{13})([-]{1}[A-Z]{1}){0,1}

It works correctly, but I'm having an issue validating that, when the duplicate is entered, its completed and correctly

/([a-zA-Z0-9]{13})([-]{1}[A-Z]{1}){0,1}/.test("ABCDEFGKLM123") // true
/([a-zA-Z0-9]{13})([-]{1}[A-Z]{1}){0,1}/.test("ABCDEFGKLM123-A") // true
/([a-zA-Z0-9]{13})([-]{1}[A-Z]{1}){0,1}/.test("ABCDEFGKLM123-AA") // true. Should return false!
/([a-zA-Z0-9]{13})([-]{1}[A-Z]{1}){0,1}/.test("ABCDEFGKLM123-") // true. Should return false!
/([a-zA-Z0-9]{13})([-]{1}[A-Z]{1}){0,1}/.test("ABCDEFGKLM123A") // true. Should return false!

I'm having problems getting this last 3 scenarios to be validated correctly

I'm still trying to figure out how to make that, if the optional part is present, its complete and correct over here https://regex101.com/r/wU4tQ5/3

Upvotes: 0

Views: 34

Answers (2)

Ammadu
Ammadu

Reputation: 1675

Try this regex

(^[a-zA-Z0-9]{13}(?:[-]{1}[A-Z]{1}){0,1}$)

Upvotes: 1

Kasravnd
Kasravnd

Reputation: 107287

There is some problem here.First you don't need {0,1} for an optional case you can simply use ? and [-]{1} can be - also [A-Z]{1} to [A-Z].Also you need to use start and end anchors to match your string from start to end to refuse of matching some string like ABCDEFGKLM123-AA

^([a-zA-Z0-9]{13})(-[A-Z])?$

See demo https://regex101.com/r/kD2wN2/1

Upvotes: 3

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