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Java Java Basics Getting Started with Java Strings, Variables, and Formatting

Challenge task 3 of 3 Strings

java.util.MissingFormatArgumentException: Format specifier '%s' ()

Can someone explain what this error means please?

Name.java
// I have setup a java.io.Console object for you named console
String firstName = "Leroy";
console.printf("%s firstName can code in Java!");

1 Answer

Ken Alger
STAFF
Ken Alger
Treehouse Teacher

Leroy;

You are close, but you aren't correctly passing the firstName variable into the String Formatter. That's what the error message is essentially saying.

Recall that if you use the %s in the string as a variable placeholder you need to pass in that variable to the printf method. That syntax looks like:

String favoriteDrink = "chocolate milk";
console.printf("My favorite drink is: %s", favoriteDrink);

Post back if you have further questions.

Ken