Regex not beginning with number

Question

How do I create a regex that matches all alphanumerics without a number at the beginning?

Right now I have "^[0-9][a-zA-Z0-9_]"

For example, 1ab would not match, ab1 would match, 1_bc would not match, bc_1 would match.

Answer 1

There are three things wrong with what you've written.

First, to negate a character class, you put the ^ inside the brackets, not before them. ^[0-9] means "any digit, at the start of the string"; [^0-9] means "anything except a digit".

Second, [^0-9] will match anything that isn't a digit, not just letters and underscores. You really want to say that the first character "is not a digit, but is a digit, letter, or underscore", right? While it isn't impossible to say that, it's a lot easier to just merge that into "is a letter or underscore".

Also, you forgot to repeat the last character set. As-is, you're matching exactly two characters, so b1 will work, but b12 will not.

So:

[a-zA-Z_][a-zA-Z0-9_]*

Debuggex Demo

In others words: one letter or underscore, followed by zero or more letters, digits, or underscores.

I'm not entirely sure this is what you actually want, at least if the regex is your whole parser. For example, in foo-bar, do you want the bar to get matched? If so, in 123spam, do you want the spam to get matched? But it's what you were trying to write.