How can I parse String to int with the default value? [duplicate]

Answer 1

You're most likely using apache.commons.lang3 already:

NumberUtils.toInt(str, 0);

Answer 2

You can use this way with String::matches like this :

String userid = "user001";
int int_userid = userid.matches("\\d+") ? Integer.parseInt(userid) : 0;

You ca also use -?\d+ for both positive and negative values :

int int_userid = userid.matches("-?\\d+") ? Integer.parseInt(userid) : 0;

Answer 3

You can try this method with a regular expression.

public static int parseWithDefault(String s, int defaultVal) {
    return s.matches("-?\\d+") ? Integer.parseInt(s) : defaultVal;   
}

Answer 4

int int_userid;
try {
    int_userid = Integer.parseInt(userid); // try to parse the given user id
} catch (Exception e) {   // catch if any exception
    int_userid = 0; // if exception assign a default value
}

Answer 5

It might be a little over-engineering, but you can use Guava's Ints.tryParse(String) with Java 8's Optionals like this:

int userId = Optional.ofNullable(Ints.tryParse("userid001")).orElse(0)

Answer 6

I believe that you can achieve what you want by the following method:

public static int parseIntWithDefault(String s, int default) {
    try {
        return Integer.parseInt(s);
    } catch(NumberFormatException e) {
        return default;
    }
}

and now just assign:

int int_userid = parseIntWithDefault(userId, 0);

Please have in mind, that using Java one should use Java good practices about formatting the code. int_userid is definitely something to improve.