JavaScript regular expression - two [a-z] followed by three [0-9] only

Question

I've got a simple regular expression:

[A-z]{2}[0-9]{3})$/g inside the following:

regForm.submit(function(){
  if ($.trim($('#new-usr').val()).match(/([A-z]{2}[0-9]{3})$/g)) {
    alert('No');
    return false;
  }
});

This is correctly reading that something like 'ab123' gives an alert and 'ab1234' doesn't. However, 'abc123' is still throwing the alert. I need it so it's only throwing the alert when it's just 2 letters followed by three numbers.

Answer 1

Try /^[A-z]{2}[0-9]{3}$/g instead.

You need to specify that the whole string needs to be matched. Otherwise you get the highlighted part matched: abc123.

(I omitted the ()'s, because you don't really need the group.)

BTW, are you sure that you want [A-z] and not just [A-Za-z]?

Answer 2

The character class [A-z] is probably not what you need.

Why?

The character class [A-z] matches some non-alphabetical characters like [, ] among others.

JS fiddle link to prove this.

This W3school tutorial recommends it incorrectly.

If you need only lowercase letters use [a-z]
If you need only uppercase letters use [A-Z]
If you need both use: [a-zA-Z]

If you want to match a string if it has 2 letters followed by 3 digits anywhere in the string, just remove the end anchor $ from your pattern:

[a-z]{2}[0-9]{3}

If you want to match a string if it has 2 letters followed by 3 digits and nothing else use both start anchor ^ and end anchor $ as

^[a-z]{2}[0-9]{3}$

Answer 3

Alternatively you can use:

/\b([A-z]{2}[0-9]{3})\b/g

if your string contains multiple words and you are trying to match one word.