Welcome to the Treehouse Community
The Treehouse Community is a meeting place for developers, designers, and programmers of all backgrounds and skill levels to get support. Collaborate here on code errors or bugs that you need feedback on, or asking for an extra set of eyes on your latest project. Join thousands of Treehouse students and alumni in the community today. (Note: Only Treehouse students can comment or ask questions, but non-students are welcome to browse our conversations.)
Looking to learn something new?
Treehouse offers a seven day free trial for new students. Get access to thousands of hours of content and a supportive community. Start your free trial today.

Jan Lundeen
5,630 PointsGetting java.util.MissingFormatArgumentException: Format specifier '%s' () error-Challenge 3 of 3-strings variables-Java
I'm getting a java.util.MissingFormatArgumentException: Format specifier '%s' () error-Challenge 3 of 3 for strings and variables in Java Basics. Challenge 3 of 3 - Now replace <YOUR NAME> in the console.printf expression with the firstName variable using the string formatter:
Here's my code:
String firstName = "Jan";
console.printf ("%s can code Java!");
Here's the error I received:
Bummer: java.util.MissingFormatArgumentException: Format specifier '%s' ()
Is "using the string formatter" referring to %s? I'm not sure what I did wrong.
Thanks,
Jan
String firstName = "Jan";
console.printf ("%s can code in Java!");// I have setup a java.io.Console object for you named console
2 Answers

Juan Jaramillo
5,481 PointsSince you used %s you have to specify which string is going to replace that %s once the console prints. It should be among the lines of :
console.printf ("%s can code Java!" , firstName);

Jan Lundeen
5,630 PointsThanks Juan! That worked.