What is a regex for Twitter-like names?

Question

I have been coding for a while but never had the need for regular expressions until recently. I need to do a regular expression that accepts usernames as Twitter does. Basically, I want to allow one underscore at a time. There can be more than one underscore in a name but these should not be consecutive characters. Alphanumeric characters are also allowed. But numbers cannot start a name.

Names such as

• _myname67
• myname67
• my_name
• _my_67_name_

are valid but

• 94myname
• __myname
• my__name
• my name

are not valid.

I have played with Rubular and come up with a couple regexes:

• /^[^0-9\s+](_?[a-z0-9]+_?)+$/i
• /^([a-z_?])+$/i

The problem I keep running into is that these match more than one underscores.

Answer 1

Edited

a = %w[
    _myname67
    myname67
    my_name
    _my_67_name_
    94myname
    __myname
    my__name
    my\ name
    m_yname
]

p a.select{|name| name =~ /\A_?[a-z]_?(?:[a-z0-9]_?)*\z/i}
# => ["_myname67", "myname67", "my_name", "_my_67_name_", "m_yname"]

You should use ( ) only for substrings that you want to capture. (?: ) is used for groupings that you do not want to capture. It is a good practice to use it whenever you do not need to refer particularly to that substring. It also makes the regex run faster.

Answer 2

Try the following ^([a-zA-Z](_?[a-zA-Z0-9]+)*_?|_([a-zA-Z0-9]+_?)*)$

I've separated two cases: the word starts with a letter, and starts with an underscore. If you don't want to allow names consisting of one symbol only replace the * with +.

maerics's solution has one problem, it doesn't capture names that have _ on the second place, such as m_yname

Answer 3

Some things are really hard to express using only regular expressions, and are generally write-only (that is, there's no way to read and understand them lately). You can use a simpler regexp (like the two ones you managed to write) and check for double underscores in your Ruby code. It doesn't hurt:

if username =~ /^[^0-9](_?[a-z0-9]+_?)+$/i and username.count('__') == 0 then ...