Welcome to the Treehouse Community

Want to collaborate on code errors? Have bugs you need feedback on? Looking for an extra set of eyes on your latest project? Get support with fellow developers, designers, and programmers of all backgrounds and skill levels here with the Treehouse Community! While you're at it, check out some resources Treehouse students have shared here.

Looking to learn something new?

Treehouse offers a seven day free trial for new students. Get access to thousands of hours of content and join thousands of Treehouse students and alumni in the community today.

Start your free trial

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

Java Basics what did I do wrong?

I don't know where I messed up on this

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

2 Answers

Ryan Ruscett
Ryan Ruscett
23,309 Points

Hey,

You have a percent s %s to be substituted as a String. Except you don't tell it what String to substitute it for.

String firstName = "Jeremiah";
console.printf("%s can code in java", firstName);

You need the comma then the variable name of the String to substitute for the %s.

let me know if this solves your problem. Thanks!

Ahhh okay now I remember haha. Thanks for your help Ryan!