NumberFormatException: For input string: ""?

Question

Code

int weight = 0;
do {
    System.out.print("Weight (lb): ");
    weight = Integer.parseInt(console.nextLine());
    if (weight <= 0) {
        throw new IllegalArgumentException("Invalid weight.");
    }
} while (weight <= 0);

Traceback

Weight (lb): Exception in thread "main" java.lang.NumberFormatException: For input string: ""
    at java.base/java.lang.NumberFormatException.forInputString(NumberFormatException.java:65)
    at java.base/java.lang.Integer.parseInt(Integer.java:662)
    at java.base/java.lang.Integer.parseInt(Integer.java:770)
    at HealthPlan.main(HealthPlan.java:46)

When I run my program, I get this exception. How do I handle it?

I want to input an integer as a weight value. I also have to use an integer value for height, but my program asks for input that are booleans and characters as well.

Someone suggested that I should use Integer.parseInt.

If I need to post more code, I'd be happy to do so.

Solution

Sometimes it simply means that you're passing an empty string into Integer.parseInt():

String a = "";
int i = Integer.parseInt(a);

Comments

[deHaar]

Make sure the console.nextLine() is not an empty String, because the line Exception in thread "main" java.lang.NumberFormatException: For input string: "" says it is...

[assylias]

Note : assuming console is a Scanner, you could use console.nextInt().

[vincrichaud]

Surround your call to parseInt() with a try-catch block catching this exception

[soufrk]

Check if console.nextLine().trim().isEmpty() is true or not.

[Jaroslaw Pawlak]

By the way, why the loop? This loop will never execute more than once...

[Shilpa Kancharla]

@assylias I tried that, and it works. Thank you!

[vincrichaud]

You can cast String to Integer, it's just that you String need to be well formatted. For example Integer.parseInt("124") is totally correct.