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

Enyang Mercy
PLUS
Enyang Mercy
Courses Plus Student 2,339 Points

i can't run or compile my code . was it just to define? the challenge was "define string variable named firstName that

// I have setup a java.io.Console object for you named console public class Introductions public static void main (String[]args) { Console console = System.console(); //Welcome to my introduction program!Check code below here. String firstName = "Mercy"; Console.printf("Hello, my name is Mercy\n"); Console.printf("Mercy is learning how to write Java\n"); } }

Name.java
// I have setup a java.io.Console object for you named console
  public class Introductions
  public static void main (String[]args) {
    Console console = System.console();
    //Welcome to my introduction program!Check code below here.
    String firstName = "Mercy";
    Console.printf("Hello, my name is Mercy\n");
    Console.printf("Mercy is learning how to write Java\n");
          }
       }

1 Answer

Samuel Ferree
Samuel Ferree
31,722 Points

you're missing an open curly on your class definition.

See below:

// I have setup a java.io.Console object for you named console
public class Introductions { // <--- This curly brace is missing
  public static void main (String[]args) {
    Console console = System.console();
    //Welcome to my introduction program!Check code below here.
    String firstName = "Mercy";
    Console.printf("Hello, my name is Mercy\n");
    Console.printf("Mercy is learning how to write Java\n");
  }
}