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

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 37,746 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

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