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Jan Lundeen
5,630 PointsGetting java.util.MissingFormatArgumentException: Format specifier '%s' () error-Challenge 3 of 3-strings variables-Java
I'm getting a java.util.MissingFormatArgumentException: Format specifier '%s' () error-Challenge 3 of 3 for strings and variables in Java Basics. Challenge 3 of 3 - Now replace <YOUR NAME> in the console.printf expression with the firstName variable using the string formatter:
Here's my code:
String firstName = "Jan";
console.printf ("%s can code Java!");
Here's the error I received:
Bummer: java.util.MissingFormatArgumentException: Format specifier '%s' ()
Is "using the string formatter" referring to %s? I'm not sure what I did wrong.
Thanks,
Jan
String firstName = "Jan";
console.printf ("%s can code in Java!");// I have setup a java.io.Console object for you named console
2 Answers
Juan Jaramillo
5,481 PointsSince you used %s you have to specify which string is going to replace that %s once the console prints. It should be among the lines of :
console.printf ("%s can code Java!" , firstName);
Jan Lundeen
5,630 PointsThanks Juan! That worked.