check if string is integer java

Trying to determine if a string is an integer

stackoverflow.com › questions › 40519580 › trying-to-determine-if-a-string-is-an-integer

You could use a regular expression.

return str.matches("\\d+");

won't work for negative numbers, though.

You could also use Integer.parseInt and catch the NumberFormatException and return true/false accordingly.

Or, you should not break the first digit you find, as you need to check all characters, only until you find one that is not a digit. Again, this does not capture negative numbers

public boolean isInteger(String str) {
    if(str == null || str.trim().isEmpty()) {
        return false;
    }
    for (int i = 0; i < str.length(); i++) {
        if(!Character.isDigit(str.charAt(i))) {
            return false;
        } 
    }
    return true;
}

Personally, option 2 is the best, but your instructions seem to imply that you need to iterate over character values

Answer from OneCricketeer on Stack Overflow

Stack Overflow

stackoverflow.com › questions › 40519580 › trying-to-determine-if-a-string-is-an-integer

java - Trying to determine if a string is an integer - Stack Overflow

Top answer

1 of 6

You could use a regular expression.

return str.matches("\\d+");

won't work for negative numbers, though.

You could also use Integer.parseInt and catch the NumberFormatException and return true/false accordingly.

Or, you should not break the first digit you find, as you need to check all characters, only until you find one that is not a digit. Again, this does not capture negative numbers

public boolean isInteger(String str) {
    if(str == null || str.trim().isEmpty()) {
        return false;
    }
    for (int i = 0; i < str.length(); i++) {
        if(!Character.isDigit(str.charAt(i))) {
            return false;
        } 
    }
    return true;
}

Personally, option 2 is the best, but your instructions seem to imply that you need to iterate over character values

2 of 6

Just check if Integer.parseInt throws exception or not.

try {
    Integer.parseInt(string);
} catch (NumberFormatException e) {
    return false;
}
return true;

Stack Overflow

stackoverflow.com › questions › 5439529 › determine-if-a-string-is-an-integer-in-java

Determine if a String is an Integer in Java - Stack Overflow

Top answer

1 of 9

364

The most naive way would be to iterate over the String and make sure all the elements are valid digits for the given radix. This is about as efficient as it could possibly get, since you must look at each element at least once. I suppose we could micro-optimize it based on the radix, but for all intents and purposes this is as good as you can expect to get.

public static boolean isInteger(String s) {
    return isInteger(s,10);
}

public static boolean isInteger(String s, int radix) {
    if(s.isEmpty()) return false;
    for(int i = 0; i < s.length(); i++) {
        if(i == 0 && s.charAt(i) == '-') {
            if(s.length() == 1) return false;
            else continue;
        }
        if(Character.digit(s.charAt(i),radix) < 0) return false;
    }
    return true;
}

Alternatively, you can rely on the Java library to have this. It's not exception based, and will catch just about every error condition you can think of. It will be a little more expensive (you have to create a Scanner object, which in a critically-tight loop you don't want to do. But it generally shouldn't be too much more expensive, so for day-to-day operations it should be pretty reliable.

public static boolean isInteger(String s, int radix) {
    Scanner sc = new Scanner(s.trim());
    if(!sc.hasNextInt(radix)) return false;
    // we know it starts with a valid int, now make sure
    // there's nothing left!
    sc.nextInt(radix);
    return !sc.hasNext();
}

If best practices don't matter to you, or you want to troll the guy who does your code reviews, try this on for size:

public static boolean isInteger(String s) {
    try { 
        Integer.parseInt(s); 
    } catch(NumberFormatException e) { 
        return false; 
    } catch(NullPointerException e) {
        return false;
    }
    // only got here if we didn't return false
    return true;
}

2 of 9

316

It's better to use regular expression like this:

str.matches("-?\\d+");

-?     --> negative sign, could have none or one
\\d+   --> one or more digits

It is not good to use `NumberFormatException` here if you can use `if-statement` instead.

If you don't want leading zero's, you can just use the regular expression as follow:

str.matches("-?(0|[1-9]\\d*)");

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