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Java Java Basics Getting Started with Java IO

kelvin muniz - roque
kelvin muniz - roque
403 Points
Posted by kelvin muniz - roque
kelvin muniz - roque
403 Points

console.printf("First name: %s", firstName); i cant see the error. claims i forgot the firstName parameter

can anyone help me. i tried sticking to the format i am learning. i cant see where i am wrong.

IO.java 
// I have imported java.io.Console for you.  It is a variable called console.
String firstName = "Kelvin";
String lastName = "muniz";
console.readLine("lastName");
console.readLine("firstName");
console.printf("First name:  %s", firstName);

1 Answer

Steven Parker
Steven Parker
231,271 Points
Steven Parker
Steven Parker
231,271 Points

The instructions asked for the first and last names to be stored in the new variables using "console.readLine", but the code here is storing fixed string literals instead. You do have some calls to "console.readLine", but the result is not being stored anywhere.

Apparently the challenge got confused and passed your previous tasks anyway, but on this one it's actually looking for the input it provides to be there and not finding it.

You'll need to start over and redo tasks 1 and 2 without using the string literals.

kelvin muniz - roque
kelvin muniz - roque
403 Points
kelvin muniz - roque
kelvin muniz - roque
403 Points

i think i understand and i have another solution that led me to the same error i am having a. here is my new code that seems better but still incorrect.

String firstName = "kelvin"; console.readLine("%s", firstName); String lastName = "Muniz"; console.readLine("%s", lastName); console.printf("First name: %s" , firstName)

i might be misunderstanding the video tutorial that teaches this challenge.

Steven Parker
Steven Parker
231,271 Points
Steven Parker
Steven Parker
231,271 Points

You still need to get rid of the string literals, and assign the result of calling the method. Also, the "readLine" method does not take any parameters.

So the first line should look like this:

String firstName = console.readLine();