Proper way to avoid parseInt throwing a NumberFormatException for input string: “”

Question

When I run parseInt:

Integer.parseInt(myString);

it throws:

NumberFormatException: For input string: \"\"

Answer 1

If the string can be empty I do it this way:

Integer.parseInt("0" + inputString)

When I'm not sure it contains only digits:

Integer.parseInt(0 + inputString.replaceAll("\\D+",""))