Validate Email with regular expression where the number of extensions in the domain part can vary

Question

I need a regular expression to match a pattern like this. I have tried using this -

if(str.match(/^([a-zA-Z-\.]+)@([a-zA-Z]{1,5})\.([a-zA-Z]{1,5})(\.[a-zA-Z]{1,5})?(\.[a-zA-Z]{2,3})?$/)){
  console.log("the email is valid");
}

But this counts user@aaa.bb.cccc as valid whereas it should not be.

The full requirements of the email to be valid -

1. The username part of the email can have only words(case insensitive), dots(.) and dashes(-). Length of 1 or more characters.

2. The domain name part can have 2 to 4 extensions. For example, username@ext.ext.ext.ext this email has 4 extensions.

3. All the extensions can only have words(case insensitive), and the length can of 1 to 5 characters long except the last extension. The last extension can only be of 2 to 3 characters long. For Example,

   user@aaa.bb.cccc.dd is valid

   user@aaa.bv.cccc is not valid

   I am having trouble with solving this part. As the extension count can vary I don't know How to check if only the last part of the email has 2 to 3 characters?

I know this type of question might be answered before. I am new to Regular Expressions and I could not find a solution to my problem.

Answer 1

You could match a single [a-zA-Z]{1,5} after the @.

Then match 0 to 2 times \.[a-zA-Z]{1,5} and match \.[a-zA-Z]{2,3} at the end to match 2-4 parts in total.

^[a-zA-Z.-]+@[a-zA-Z]{1,5}(?:\.[a-zA-Z]{1,5}){0,2}\.[a-zA-Z]{2,3}$

Regex demo

If you want to match "words", you might also use \w instead of using [a-zA-Z] to also match digits and _

const pattern = /^[a-zA-Z.-]+@[a-zA-Z]{1,5}(?:\.[a-zA-Z]{1,5}){1,2}\.[a-zA-Z]{2,3}$/;
[
  "user@aaa.bb.cccc.dd",
  "user@aaa.bb.cccc.dd.ee",
  "user@aaa.bv.cccc"
].forEach(s => {
  console.log(`the email is ${pattern.test(s) ? "" : "not "}valid for ${s}`)
});