Regex To Include Alphanumeric And _
Solution 1:
Regular expressions are read as patterns which actually match characters in a string, left to right, so your pattern actually matches an alphanumeric, THEN an underscore (0 or more), THEN at least one character that is not a hyphen, dollar, or whitespace.
Since you're trying to alternate on character types, just use a character class to show what characters you're allowing:
[\w_]
This checks that ANY part of the string matches it, so let's anchor it to the beginning and and of the string:
^[\w_]$
And now we see that the character class lacks a quantifier, so we are matching on exactly ONE character. We can fix that using +
(if you want one or more characters, no empty strings) or *
(if you want to allow empty strings). I'll use +
here.
^[\w_]+$
As it turns out, the \w
character class already includes the underscore, so we can remove the redundant underscore from the pattern:
^[\w]+$
And now we have only one character in the character class, so we no longer need the character class brackets at all:
^\w+$
And that's all you need, unless I'm missing something about your requirements.
Solution 2:
Yes, you are semi-correct if the closing bracket was not escaped and you edited your regex a bit. Also the token \w
matches underscore, so you do not need to repeat this character. Your regular expression says:
\w # word characters (a-z, A-Z, 0-9, _)
_* # '_' (0or more times)
[^-$\s] # anycharacterexcept: '-', '$', whitespace (\n, \r, \t, \f, and " ")
You could simply write your entire regex as follows to match word characters:
\w+ # word characters ( a-z, A-Z, 0-9, _ ) (1 or more times)
If you want to match an entire string, be sure to anchor your expression.
^\w+$
Explanation:
^ # the beginning of the string
\w+ # word characters ( a-z, A-Z, 0-9, _ ) (1 or more times)$ # before an optional \n, and the end of the string
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