Regex to validate an alphanumeric string with optional space

Question

I have been working on a regex for validating an alphanumeric string with the rules as below:

The first FOUR starting characters must be numbers and last TWO characters must be alphabets.
The space is OPTIONAL but must be placed between two characters, meaning trailing space is not allowed.
The length of postal code must be 6 characters if SPACE is not included and 7 characters if space is included.

Eg.

1111 ZZ
111 1ZZ
1 111ZZ
1111ZZ

I tried using ^[0-9]{4}[A-Za-z]{2}$|^(?=[\d|\D]+ [\d|\D]+).{7}$ but this also validates 9999 1A as TRUE which should actually be FALSE.

Any leads or help will be appreciated :)

Is there any particular reason why you can't remove all spaces (leading, middle or trailing), and still have the same semantic 'value'? — TreffnonX
– TreffnonX, Commented Oct 20, 2020 at 12:27

malarres · Accepted Answer · 2020-10-21 07:53:25Z

1

(?=^.{6,7}$)^(([0-9] ?){4}( ?[a-zA-Z]){2})$

will match

1111 ZZ
111 1ZZ
1 111ZZ
1111ZZ
1111 ZZ

but not

9999 1A
11111 Z
1111111
11 11 ZZ

https://regex101.com/r/lByOx6/1

EDIT: explanation

The "Positive Lookahead" part:

(?=^.{6,7}$) this only matches if the string meets the requirements, BUT it does not 'consume' the characters.
- . is any character
- {6,7} is about repetitions

so (?=^.{6,7}$) is matched if the string has 6 or 7 characters, no matter what

Then the following part already 'consumes' the string to say that I want at the start 4 repetitions of numbers and optionally space, and at the end 2 repetitions of letters and optionally space. The second part would accept strings such as 1 1 1 1 Z Z but as those are more than 7 characters, the first part wouldn't let the string match.

edited Oct 21, 2020 at 7:53

answered Oct 20, 2020 at 12:40

malarres

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1 Comment

Sarthak Mehra Over a year ago

Thanks! that solves the problem. Can you also explain a little bit how the regex is working? It will help :)

TreffnonX · Accepted Answer · 2020-10-20 12:36:53Z

1

I suggest simplifying the problem ahead of time, by reducing all white spaces, which you seem to be uninterested in anyway:

var candidate = input.replaceAll(/\s/mg, '');

Then the regex is simply: /^\d{4}[A-Za-z]{2}$/

If, however, you need to validate, that there actually are no leading or trailing white spaces, you can validate that ahead of time, and return a negative result right away.

edited Oct 20, 2020 at 12:36

answered Oct 20, 2020 at 12:31

TreffnonX

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2 Comments

Max Xapi Over a year ago

Once you've removed the spaces, [0-9A-Za-z]* becomes wrong (it would ie validate "111111ZZ"), /^\d{4}[A-Za-z]{2}$/ is enough

TreffnonX Over a year ago

@MaxXapi You are correct. I forgot about the '6 character' constraint.

The fourth bird · Accepted Answer · 2020-10-20 20:49:46Z

0

Another option is to check if the string contains an optional space between the first and the last non whitespace character.

Then match the first digit followed by 3 digits separated by an optional space and 2 or 3 times a char a-zA-Z or a space.

Using a case insensitive match:

^(?=\S+ ?\S+$)\d(?: ?\d){3}[A-Z ]{2,3}$

Explanation

^ Start of string
(?= Positive lookahead, assert what on the right is
- \S+ ?\S+$ Match optional space between the first and the last non whitespace char
) Close lookahead
\d(?: ?\d){3} Match a digit and repeat 3 times an optional space and a digit
[a-zA-Z ]{2,3} Match 2-3 times either a char a-zA-Z or a space
$ End of string

Regex demo

answered Oct 20, 2020 at 20:49

The fourth bird

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