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Java Java Basics Getting Started with Java IO

Posted by Martin Geil
Martin Geil
432 Points

Where I need to place "%s" ?

.

IO.java 
// I have imported java.io.Console for you.  It is a variable called console.
String firstName = console.readLine("Martin");
String lastName = console.readLine("Geil");
console.printf("firstName:");

1 Answer

Steve Hunter
Steve Hunter
57,712 Points

Hi Martin,

I covered this in my other answer but I thought I'd answer this too for completeness.

Focussing just on the output string here, the printf - the rest is answered elsewhere.

The printf method outputs a string. That string can be contained within double-quotes, as you have done. Anything within double quotes is just a string; it cannot access variable values. So you've got "firstName" - that's not the variable, that's just a string that says "firstName".

To access the value of a variable within a string, you can use % characters - there are % characters for numbers, strings etc. You'll cover it all in the courses. You insert the %s, for a string insertion, where you want the string to be inserted. Then, after the double quotes (and a comma) you type the variable name:

String firstName = "Steve" // I didn't ask for user input
console.printf("First name: %s", firstName); // Value Steve gets inserted where the %s is

I hope that helps,

Steve.