java - java regular expression optional characters

Question

I'm having problems getting my regular expression working.

The regular expression is:

([0-9]m)* ([0-9]f)*

A digit must come before "m" or "f" but "m" or "f" are optional. Example are:

1m 2f
1m
6f

What have I done wrong?

Answer 1

The * means match the previous token 0 or more times, which doesn't look like what you want.

These should help you to build the regular expression you need:

• ? to mean 0 or 1 matches.
• | for alternation.
• (?:...) for a non-capturing group.
• ^ and $ to anchor at the start and end of the string.

Knowing that, I imagine that you can probably find a solution by yourself, but for the sake of completion, I'll show one possible solution.

---

Your question isn't very clear so I'm just going to assume that you want the following to match:

1m 2f
1m
6f 
0m

and that you want the following to fail to match:

1
m    
11m
1m 1m
2f 3m
1m  2f
"1m 2f"

If those assumptions are incorrect, then please make your question more clear.

With those assumptions, try this:

^[0-9]m(?: [0-9]f)?$|^[0-9]f$

If you also want 2f 3m to match then use this:

^[0-9]m(?: [0-9]f)?$|^[0-9]f(?: [0-9]m)?$