Regex: Check if string contains at least one digit
I'm surprised nobody has mentioned the simplest version:
\d
This will match any digit. If your regular expression engine is Unicode-aware, this means it will match anything that's defined as a digit in any language, not just the Arabic numerals 0-9.
There's no need to put it in [
square brackets]
to define it as a character class, as one of the other answers did; \d
works fine by itself.
Since it's not anchored with ^
or $
, it will match any subset of the string, so if the string contains at least one digit, this will match.
And there's no need for the added complexity of +
, since the goal is just to determine whether there's at least one digit. If there's at least one digit, this will match; and it will do so with a minimum of overhead.
The regular expression you are looking for is simply this:
[0-9]
You do not mention what language you are using. If your regular expression evaluator forces REs to be anchored, you need this:
.*[0-9].*
Some RE engines (modern ones!) also allow you to write the first as \d
(mnemonically: digit) and the second would then become .*\d.*
.
In Java:
public boolean containsNumber(String string)
{
return string.matches(".*\\d+.*");
}