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Java Java Basics Getting Started with Java IO

Posted by Ben Cotton
Ben Cotton
Courses Plus Student 105 Points

Not sure where I'm going wrong, I thought I had to stipulate the variable's name within the brackets (or parentheses)?

It's giving me an error but I'm not sure where I've gone wrong. I'm wondering if I've made a syntax error, but since I'm very new to Java I'm not easily able to diagnose the problem.

Thank you in advance.

IO.java 
// I have imported java.io.Console for you.  It is a variable called console.
String firstName= console.readLine("What is your name?  ");
String lastName= console.readLine("What is your surname?  ");
console.printf= ("First name: %s."); firstName;

All solved now :D

I'm not sure if there's a way to change the status of this thread to 'resolved'?

Ian Holden
Ian Holden
8,556 Points
Ian Holden
Ian Holden
8,556 Points

Ah yes! I missed this in my answer haha! That certainly will be an issue too.

3 Answers

Ben Cotton
Ben Cotton
Courses Plus Student 105 Points

Wait I think I've figured it. I think it should be "console.printf(" rather than "console.printf=".

Ian Holden
Ian Holden
8,556 Points
Ian Holden
Ian Holden
8,556 Points

Ok, I am not an expert with this method but I have noticed an issue with your final line of code, so I hope this helps in some way.

This line: console.printf= ("First name: %s."); firstName;

will print the string that you have written: "First name: %s.". However, I can see that you want to print the firstName variable which at the moment is sitting on its own as an individual statement. It is an individual statement at the moment because you have a semi-colon at the end of the closing brackets after your string in the line above. This means that you have something that in the compiler will be read like this:

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

Now you can see that you actually have two statements, the first that may print your string and the second which will do nothing. I think the second line could be causing your error! Currently, the variable is just being named but not used in the statement which makes it invalid.

The next few lines are just a guess but could you try using the firstName variable in the first line statement like so:

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

See how this goes and let me know :) I hope this worked/helped.

Ben Cotton
Ben Cotton
Courses Plus Student 105 Points

Thanks for the response, Ian :D

I think I messed up the last line trying to find a solution to my problem (which was the '( or =' conundrum above) and therefore caused myself more hassle haha!

All resolved now, though, lessons learnt.

Thanks for the help :)

Johan Prieto
Johan Prieto
6,513 Points
Johan Prieto
Johan Prieto
6,513 Points

Yeah Ben, it has to be console.printf("") like you said :D

Ben Cotton
Ben Cotton
Courses Plus Student 105 Points

Thanks very much for the response, Johan. I got it working :D