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Java Java Basics Getting Started with Java IO

Posted by Gustavo Borges
Gustavo Borges
309 Points

I cant pass this exercise!

In my perspective, I'm not doing anything wrong.

But this message keeps coming, every time I ask to check my work: "Did you forget to pass the lastName parameter to the printf function?"

I think I a;ready passed the lastName parameter to the printf function, right?

Could someone help me?!

IO.java 
// I have imported java.io.Console for you.  It is a variable called console.
String firstName = console.readLine("What is your first name?  ");
String lastName = console.readLine("What is your last name?  ");

console.printf("First name", firstName); 
console.printf("%s", firstName);

console.printf("Last name: ", lastName); 
console.printf("%s", lastName);
Grigorij Schleifer
Grigorij Schleifer
10,365 Points

Your code is almost fine, but the environment is complaining because two lines are missing the string formatters.

3 Answers

Gustavo Borges
Gustavo Borges
309 Points
Gustavo Borges
Gustavo Borges
309 Points

Thanks a lot!

That's weird... because before the lastName step, I used two lines (as you could in my code) and it worked.

Once I tried to do the same thing in the last step, the system didn't allow me to finish it.

But I understand your point!

Again, thanks a lot!!!

Grigorij Schleifer
Grigorij Schleifer
10,365 Points

Hey,

glad I could help :)

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

Hey Gustavo,

look at my code suggestion. I added just a few modifications.

String firstName = console.readLine("What is your first name?  ");
String lastName = console.readLine("What is your last name?  ");

console.printf("First name: %s", firstName); 
console.printf("Last name: %s", lastName);

Grigorij

David Lacedonia
David Lacedonia
13,627 Points
David Lacedonia
David Lacedonia
13,627 Points

This works:

console.printf("First name: %s", firstName);
console.printf("Last name: %s", lastName);

Don't pass any parameter when you dont have %s.

If you want the printf methods in diferent lines, also you can do...

console.printf("First name: ");
console.printf("%s",firstName);
console.printf("Last name: ");
console.printf("%s",lastName);