How to detect exact length in regex

Question

I have two regular expressions that validate the values entered.

One that allows any length of Alpha-Numeric value:

@\"^\\s*(?[A-Z0-9]+)         


        

Answer 1

I think what you're trying to say is that you don't want to allow any more than 10 digits. So, just add a $ at the end to specify the end of the regex.

Example: @"^\s*(?[0-9]{10})$"

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Here's my original answer, but I think I read you too exact.

string myRegexString = `@"(?!(^\d{11}$)` ... your regex here ... )";

That reads "while ahead is not, start, 11 digits, end"

Answer 2

If it's single line, you could specify that your match must happen at the end of the line, like this in .net ...

^\s*([0-9]{10})\z

That will accept 1234567890 but reject 12345678901.

Answer 3

If you are trying to match only numbers that are 10 digits long, just add a trailing anchor using $, like this:

^\s*(?:[0-9]{10})\s*$

That will match any number that is exactly 10 digits long (with optional space on either side).

Answer 4

This should match only 10 digits and allow arbitrary numbers of whitespaces before and after the digits.

Non-capturing version: (only matches, the matched digits are not stored)

^\s*(?:\d{10})\s*$

Capturing version: (the matched digits are available in subgroup 1, as $1 or \1)

^\s*(\d{10})\s*$

Answer 5

var pattern =/\b[0-9]{10}$\b/;

// the b modifier is used for boundary and $ is used for exact length

Answer 6

You could try alternation?

^\s*(?\d{1,10}|\d{12,})