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

Robert Brasso
Robert Brasso
1,277 Points

My code works but it won't let me move on. I've tried a bunch of variations that work, but it won't let me proceed.


// I have setup a java.io.Console object for you named console
String firstName = "Nick";
console.printf (firstName);

1 Answer

Rob Bridges
.a{fill-rule:evenodd;}techdegree seal-36
Rob Bridges
Full Stack JavaScript Techdegree Graduate 35,467 Points

Hey Chris, Two things. The 1st and most simpelest is the string that they are looking for is "<Your name> can code in java!"

You're just printing off the variable without the code.

The other thing, that I think gets to the heart of the challenge is you need to actually pass in the variable as a placeholder at first.

It would look something like this.

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

I'd go back and review the video a bit to see what the printf function can do for you.

Thanks, hope this helped.