Regular expression to check if a String is a positive natural number

问题

I want to check if a string is a positive natural number but I don't want to use Integer.parseInt() because the user may enter a number larger than an int. Instead I would prefer to use a regex to return false if a numeric String contains all "0" characters.

if(val.matches("[0-9]+")){
    // We know that it will be a number, but what if it is "000"?
    // what should I change to make sure 
    // "At Least 1 character in the String is from 1-9"
}

Note: the string must contain only 0-9 and it must not contain all 0s; in other words it must have at least 1 character in [1-9].

回答1:

You'd be better off using BigInteger if you're trying to work with an arbitrarily large integer, however the following pattern should match a series of digits containing at least one non-zero character.

\d*[1-9]\d*

Debuggex Demo

Debugex's unit tests seem a little buggy, but you can play with the pattern there. It's simple enough that it should be reasonably cross-language compatible, but in Java you'd need to escape it.

Pattern positiveNumber = Pattern.compile("\\d*[1-9]\\d*");

Note the above (intentionally) matches strings we wouldn't normally consider "positive natural numbers", as a valid string can start with one or more 0s, e.g. 000123. If you don't want to match such strings, you can simplify the pattern further.

[1-9]\d*

Debuggex Demo

Pattern exactPositiveNumber = Pattern.compile("[1-9]\\d*");

回答2:

If you want to match positive natural numbers, written in the standard way, without a leading zero, the regular expression you want is

[1-9]\d*

which matches any string of characters consisting only of digits, where the first digit is not zero. Don't forget to double the backslash ("[1-9]\\d*") if you write it as a Java String literal.

来源：https://stackoverflow.com/questions/24645577/regular-expression-to-check-if-a-string-is-a-positive-natural-number

标签

java

regex