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

I have created my class as below public class Name{ Why am I getting error message saying this is incorrect?

Please let me know Thanks

Name.java
// I have setup a java.io.Console object for you named con
public class Name{
  public static void main(String args[]){
    String firstName ="Gary";
  }
  {
    Console console = System.console();
console.printf("Gary, can code in Java!");
  }
}
Jeff Janes
Jeff Janes
8,033 Points

It looks like you're ending your class (with the }s) before the Console line. See if the code below works...

// I have setup a java.io.Console object for you named con
public class Name{
  public static void main(String args[]){
    String firstName ="Gary";
  }
    Console console = System.console();
    console.printf("Gary, can code in Java!");
}

3 Answers

Hi Jeff James,

no still getting the same error

JavaTester.java:53: error: illegal start of expression public class Name{ ^ JavaTester.java:58: error: expected console.printf("Gary can code in Java!"); ^ JavaTester.java:58: error: illegal start of type console.printf("Gary can code in Java!"); ^ 3 errors

Jeff Janes
Jeff Janes
8,033 Points

Actually, after looking at this challenge, the answer is a lot more simple than you think. Can you try the code below? It seems you're missing the part with setting the String to appear within the line.

%s will be replaced with whatever String you call after the comma in the printf method.

String firstName = "Gary";

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

As Jeff Janes has said, you don't need the class or the Console console = System.Console(); part, since Craig has already done them.

brian colliflower
brian colliflower
15,433 Points

I agree with Isaiah and Jeff you have a lot of extra code you do not need you are getting tripped up on the curly braces and the extra code the whole code to finish it should look like this String firstName = "Brian"; console.printf("%s can code in Java!", firstName);