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

Joshua Lowell
Joshua Lowell
9,088 Points
Posted by Joshua Lowell
Joshua Lowell
9,088 Points

I replaced my name with %s and put , firstName"); at the end of the print expression and it still has an error.

Wouldn't it be console.printf ("%s can code in Java!, firstName");

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

2 Answers

Grigorij Schleifer
Grigorij Schleifer
10,365 Points
Grigorij Schleifer
Grigorij Schleifer
10,365 Points

Hi Josh,

it is a bug inside the envirenment.

Remove the space between printf and the parenthesis. And move the last " sign after Java!

Like this:

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

Grigorij

Derek Markman
Derek Markman
16,291 Points

Just to add a tiny bit of clarification, the spacing between "printf" and the opening parenthesis is an environment bug. But as far as his quotation marks are concerned, that's a syntax error. Its important he understands that. :)

Grigorij Schleifer
Grigorij Schleifer
10,365 Points
Grigorij Schleifer
Grigorij Schleifer
10,365 Points

Hey Derek,

you absolutely right!!! Thank you for completing :thumbsup:

Grigorij Schleifer
Grigorij Schleifer
10,365 Points
Grigorij Schleifer
Grigorij Schleifer
10,365 Points

Hey Josh,

you are welcome. See you in the forum

Grigorij

Bertan Ulusoy
Bertan Ulusoy
18,047 Points
Bertan Ulusoy
Bertan Ulusoy
18,047 Points

Hi, You need to put your quote before variable name in console printf method. Just like this. You know, %s refers that firstName string variable.

String firstName = "Joshua Lowell"; console.printf ("%s can code in JAVA!", firstName);