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

Posted by joseph morgan
joseph morgan
314 Points

I got an error I do not understand.

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

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

1 Answer

Mark Sebeck
MOD
Mark Sebeck
Treehouse Moderator 38,297 Points
Mark Sebeck
Mark Sebeck
Treehouse Moderator 38,297 Points

hi joseph morgan . So your error is a Missing Format Argument. You have the Format Specifier (%s) but remember the Format Argument should be the second parameter after the string. Having firstName in quotes will just print the string firstName instead of the value stored in the variable. 

console.printf("%s can code in Java!", firstName);

Hope this helps and keep at it!

Grigorij Schleifer
Grigorij Schleifer
10,365 Points
Grigorij Schleifer
Grigorij Schleifer
10,365 Points

This is very useful and honestly better than the println option IMHO.