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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?

// 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
Ken Alger
Treehouse Teacher


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.