Checking if a string parameter contains only digits

There is one major flaw that will come back and haunt you with this code. If you enter an invalid number, there is absolutely no indication whatsoever that it was invalid.

If it is not valid, you should do this:

throw new IllegalArgumentException("Invalid mobile number: " + mobile);

Incorporating Heslacher's suggestions I would do this:

public void setMobile(String mobile) {
    if (!containsOnlyDigits(mobile)) {
        throw new IllegalArgumentException("Invalid mobile number: " + mobile);
    }
    this.mobile = mobile;
}

You should extract the validation, meaning the checking if the string only contains digits, to a separate method like so

private static boolean containsOnlyDigits(final String value) {
    for (int i = 0; i < value.length(); i++) {
        if(!Character.isDigit(value.charAt(i))) {
            return false;
        }
    }
    return true;
}  

and use it in the setMobile() method like so

public void setMobile(String mobile) {
    if(containsOnlyDigits(mobile)) {
        this.mobile = mobile;
    }
}  

So each method is doing only one thing.

Because the method setMobile() is public you should check if the passed mobile is not null.

Disclaimer: Written in the editor without testing.

You can use regex ^\d{1,10}$ to check this as below example:

private static final String regex="^\\d{1,10}$";

public static void main(String[] args) {
    System.out.println("123456".matches(regex));
    System.out.println("123abc".matches(regex));
}

Where regex means:

• ^ -> starts of string
• \d -> only digits
• {1, 10} -> min 1 digit, max 10 digits
• $ -> end of string

So, complete regex -> input string starts with digit, should be minimum 1 digit, max 10 digits, and also ends with digit. And no other characters allowed in between.