Check input string to int

Question

I have this method:

public static int parseInt(String str) {
    if (isValidNumber(str)) {
        int sum = 0;
        int position = 1;
        for (int i = str.length() - 1; i >= 0; i--) {
            int number = str.charAt(i) - '0';
            sum += number * position;
            position = position * 10;
        }
        return sum;
    }
    return -1;
}

which converts a string into a integer. And as you can see it is (at the moment) in a if-statement with a method which checks if the input is a valid input for my purpose:

public static boolean isValidNumber(String str) {
    for (int i = 0; i < str.length(); i++) {
        char c = str.charAt(i);
        if(c >= '0' && c <= '9'){
            return true;
        }
    }
    return false;
}

I want the string to be number only (negative and positive) no other is allowed. At that time a string i.e 1a1a will be converted to a integer which it shouldn't whereas -1 will not be converted. I think you guys understand what I mean. I don't know how to do that.

Please help!

isValidNumber() is wrong, because if the first number is a digit, then it returns true, even if the rest are not digits. Instead do c < '0' || c > '9' and switch the places of return true and return false. — RaminS
– RaminS, Commented Dec 10, 2016 at 19:04
Oh thats cool, thanks! But can you also help me to get negativ numbers working? I have no idea how to make a exception for the char "-" — ethanqt
– ethanqt, Commented Dec 10, 2016 at 19:08
Negative numbers don't work either, because '-' is not an integer. You need to do a new section in isValidNumber() for if(str.charAt(0) == '-'). Then instead of i = 0 in your for loop, start with i = 1. — RaminS
– RaminS, Commented Dec 10, 2016 at 19:09
you may ant ti use a regular exprression rather than iterating over the (possible) digits yourself. Refer Java API docs.oracle.com/javase/8/docs/api/java/lang/… docs.oracle.com/javase/7/docs/api/java/util/regex/Pattern.html — Timothy Truckle
– Timothy Truckle, Commented Dec 10, 2016 at 19:10

Dani · Accepted Answer · 2016-12-10 19:22:19Z

Try this:

CODE:

public class validNumbers {

    public static void main(String[] args) {

        System.out.println(parseInt("345"));
        System.out.println(parseInt("-345"));
        System.out.println(parseInt("a-345"));
        System.out.println(parseInt("1a5b"));
    }

    public static int parseInt(String str) {
        String numberWithoutSign = removeSign(str);
        if (isValidNumber(numberWithoutSign)) {
            int sum = 0;
            int position = 1;
            for (int i = numberWithoutSign.length() - 1; i >= 0; i--) {
                int number = numberWithoutSign.charAt(i) - '0';
                sum += number * position;
                position = position * 10;
            }
            if(isNegative(str)){
                return -(sum);
            }else{
                return sum;
            }
        }
        return -1;
    }

    /**
     * Removes sign in number if exists
     */
    public static String removeSign(String number){
        if(number.charAt(0) == '+' || number.charAt(0) == '-'){
            return number.substring(1);
        }else{
            return number;
        }
    }
    /**
     * Determines if a number is valid
     */
    public static boolean isValidNumber(String number) {
        for (int i = 0; i < number.length(); i++) {
            char c = number.charAt(i);
            if(c >= '0' && c <= '9'){
                continue;
            }else{
                return false;
            }
        }
        return true;
    }

    /**
     * Determines if a number is negative or not
     */
    public static boolean isNegative(String number){
        if(number.charAt(0) == '-'){
            return true;
        }else{
            return false;
        }
    }

}

OUTPUT:

345
-345
-1
-1

daN · Accepted Answer · 2016-12-10 19:10:13Z

0

To check if a string is a real number you can use a method like this:

    public static boolean isInteger(String str) {
        try {
            Integer.parseInt(str);
            return true;
        } catch (NumberFormatException nfe) {}
        return false;
    }

answered Dec 10, 2016 at 19:10

daN

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1 Comment

RaminS Over a year ago

He's creating his own parseInt() method, though. Using the one in the Java library is probably not an option.

Buhake Sindi · Accepted Answer · 2016-12-10 19:30:19Z

0

The problem is with your function isValidNumber. It should return a false on first occurrence of a non numeric value, as follows:

public static boolean isValidNumber(String str) {
    for (int i = 0; i < str.length(); i++) {
        char c = str.charAt(i);
        if(!(c >= '0' && c <= '9')){
            if (i > 0) {
                return false;
            }

            //This will only be invoked when `i == 0` (or less, which is impossible in this for loop), so I don't need to explicitly specify it here, as I have checked for `i > 0` in the above code...    
            if (c != '-' && c != '+') {
                return false;
            }
        }
    }

    return true;
}

edited Dec 10, 2016 at 19:30

answered Dec 10, 2016 at 19:12

Buhake Sindi

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4 Comments

OneCricketeer Over a year ago

Could Character.isDigit be used here?

Buhake Sindi Over a year ago

Absolutely. I think the OP wants to also include negative numbers as well.

RaminS Over a year ago

Your latest edit ruins everything, because now 5-6+33- returns true, does it not?

Buhake Sindi Over a year ago

@Gendarme, no since the charAt(1) == '-' and the if (i > 0) will kick in, thus returning a false.

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