I am trying to build a regular expression in javascript that checks for 3 word characters however 2 of them are are optional. So I have:

/^\w\w\w/i

what I am stumped on is how to make it that the user does not have to enter the last two letters but if they do they have to be letters

I am trying to build a regular expression in javascript that checks for 3 word characters however 2 of them are are optional. So I have:

/^\w\w\w/i

what I am stumped on is how to make it that the user does not have to enter the last two letters but if they do they have to be letters

• javascript
• regex

• Does /^\w\w?\w?$/i work for you? – tloflin Commented May 4, 2010 at 21:22
• I did have to change it to \d and \Dbecause cause of the data i was testing but the answers were still correct about the optional characters – Anthony Commented May 6, 2010 at 22:58

2 Answers 2

You can use this regular expression:

/^\w{1,3}$/i

The quantifier {1,3} means to repeat the preceding expression (\w) at least 1 and at most 3 times. Additionally, $ marks the end of the string similar to ^ for the start of the string. Note that \w does not just contain the characters a–z and their uppercase counterparts (so you don’t need to use the i modifier to make the expression case insensitive) but also the digits 0–9 and the low line character _.

Like this:

/^\w\w?\w?$/i

The ? marks the preceding expression as optional.

The $ is necessary to anchor the end of the regex.
Without the $, it would match a12, because it would only match the first character. The $ forces the regex to match the entire string.