Won't recognize %s in Java when using console.printf

Question

It's giving me the message, "Did you forget to pass the firstName parameter to the printf function?" when I try to run my code. I can't figure out what's wrong with it.
IO.java
// I have imported java.io.Console for you.  It is a variable called console.
String firstName = "";
console.readLine("First name?", firstName);
String lastName = "";
console.readLine("Last name?", lastName);
console.printf("First name: %s", firstName);

Rob Bridges · Accepted Answer

Hey Frank.
What Mark is trying to say is that you're never assigning the value of your of the input from console.readLine() into the variable, you intialize it as blank. There's no reason to do that yet unless you wanted to continue to ask while I value was not null.
To properly assign a value with the readLine fuction you'd use something like below
```java
String myVariable = console.readLine("What is your variable?");
```
Then I could use the assignment to do something below like the challenge is requesting by saying
```java
console.printf("Your variable is %s", myVariable);
```
You are running into trouble because you are trying to pass in the variable value in the same line where you try to find it. Your readLine() command does not need to have the value passed into it. This will cause the error that you are seeing. 
Thanks, let me know if this doesn't help.

mark gilchrist · Answer

That is because you are not assigining a value to firstname
change 
        String firstName = "";
        console.readLine("First name?", firstName);
to 
       String firstName = console.readLine("First name?", firstName);

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Frank Tocci

Frank Tocci

Won't recognize %s in Java when using console.printf

2 Answers

Rob Bridges

Rob Bridges

Frank Tocci

Frank Tocci

mark gilchrist

mark gilchrist

Frank Tocci

Frank Tocci