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Martin Geil
432 PointsWhere I need to place "%s" ?
.
// 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
57,712 PointsHi 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.