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

Priya Patil
Priya Patil
2,471 Points
Posted by Priya Patil
Priya Patil
2,471 Points

I wrote program as mentioned below String FirstName = "priya"; console.printf("my name is%s",FirstName); no output??

String FirstName = "priya"; console.printf("my name is%s",FirstName);

Name.java 
// I have setup a java.io.Console object for you named console
String FirstName = "priya";
console.printf("my name is%s",FirstName);

3 Answers

Michael Hess
Michael Hess
24,512 Points
Michael Hess
Michael Hess
24,512 Points

Hi Priya Patil,

Everything is looking okay except for two things:

  1. the firstName variable should be in camelCase. Camel case is a naming convention in which several words are joined together and the first letter of every word is capitalized.

  2. The challenge wants us to print "YOUR NAME can code in Java!".

The end result is:

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

If you have any other questions feel free to ask! Hope this helps!

Marcus Parsons
Marcus Parsons
15,719 Points
Marcus Parsons
Marcus Parsons
15,719 Points

Hi Priya,

firstName is case sensitive as are a lot of methods and functions in Java, so you have to change FirstName to firstName, the camel case representation. And then the 2nd string it tells you to print out is "<YOUR NAME> can code in Java!" where <YOUR NAME> should be a placeholder for your name. You wrote "My name is%s". Your code should be:

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