As I said, your regex is almost right, you just need to remove the% unnecessary%, making some corrections it would look like this:

^[A-Z]{2}[1-9]{9}[A-Z]{2}$

I did a test on this link , you can access it and test some modifications as well.

Just to complement the @Paz user response , a brief explanation of why your attempt did not work.

Brackets define a class or character set . For example, [ABC] means "the letter A or the letter B or the letter C". It is an expression that will take only one character.

There are some shortcuts like [A-Z] , which means "any letter from A to Z", but you can also put other characters together. For example, [3A-Z] means "the 3 digit or any letter from A to Z".

Within the brackets you can also place the opening bracket itself, that is, [[] means " [ " (since it is enclosed in square brackets). See:

console.log(/[[]/.test('[')); // true

Now the closing bracket should be escaped with \ , otherwise regex will understand that the first bracket is being closed:

// "]" escapado com "\"
console.log(/[\]]/.test(']')); // true

So, the first part of your regex ( [[A-Z] ) means "the character [ or a letter from A to Z":

console.log(/[[A-Z]/.test('A')); // true
console.log(/[[A-Z]/.test('[')); // true

Next, you place the% cos_de% quantifier, which will accept two occurrences of this regex, followed by 9 digits and 2 letters. And in the end, the last {2} corresponds to the ] character itself, since there is no equivalent opening bracket (all ] has already been closed).

This means that your regex will only accept strings with [ at the end (in addition to accepting ] at start):

console.log(/^[[A-Z]{2}[1-9]{9}[A-Z]{2}]$/.test('[A123456789AB]')); // true
console.log(/^[[A-Z]{2}[1-9]{9}[A-Z]{2}]$/.test('AZ123456789AB')); // false

That's why it's the right thing to remove these brackets from the beginning and the end. Another point of attention is that in JavaScript a regex is delimited by the slashes, so [ must be replaced by /$ :

console.log(/^[A-Z]{2}[1-9]{9}[A-Z]{2}$/.test('[A123456789AB]')); // false
console.log(/^[A-Z]{2}[1-9]{9}[A-Z]{2}$/.test('AZ123456789AB')); // true

Include zero or not

Another point that has not been clear is whether you need all the digits or not. I say this because $/ considers only digits 1 through 9 (ie, does not accept zero):

console.log(/[1-9]/.test('0')); // false
console.log(/[1-9]/.test('1')); // true

If you also need zero, simply change to [1-9] , or simply to [0-9] :

console.log(/[0-9]/.test('0')); // true
console.log(/\d/.test('0')); // true

That is, the expression would look like:

 // aceitar todos os dígitos (incluindo o zero)
console.log(/^[A-Z]{2}\d{9}[A-Z]{2}$/.test('AZ123456789AB')); // true
console.log(/^[A-Z]{2}\d{9}[A-Z]{2}$/.test('AZ023456789AB')); // true

For the AZ123456789AZ pattern, this is enough:

^[A-Z]{2}\d{9}[A-Z]{2}$

Explaining:

• ^ e $ , are a border that represents the beginning and end of one, are important because if a letter exceeds match will return to where it is valid, so it would not fully validate the pattern, which I explained can be seen in < a href="https://regex101.com/r/vbcOE3/1"> regex101 .
• % with% set that takes letters in the range of [A-Z] to A , this does not include lowercase.
• Z exactly 2 letters
• {2} shortcut to set \d