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

To print out name I type in %s to print out my string but it says missing format argument

Thank you for the guidance.

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

1 Answer

Jennifer Nordell
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STAFF
.a{fill-rule:evenodd;}techdegree
Jennifer Nordell
Treehouse Teacher

Hi there! You're soooo close here. We use the %s to tell Java that we're going to insert a string stored in a variable here. But then, we have to say which variable to insert in that spot. Take a look:

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

So this will use the printf method on the console class. Inside that we send a string with the %s (which we sometimes call a token) to tell it to insert the string value assigned to a variable in this spot. Then we use a comma and list the variable name that holds the string to be inserted. It's also important to note that your string was missing the word "in" which I have corrected here.

Hope this helps! :sparkles: