User Input not working with keyboard.nextLine() and String (Java)

Question

I recently started learning java during my spare time. So to practice, I\'m making a program that takes a temperature (Celsius or Fahrenheit) and converts it to the opposite

Answer 1

Never use Scanner#nextLine after Scanner#nextInt. Whenever you hit enter button after Scanner#nextInt than it will skip the Scanner#nextLine command. So, Change from

 int temp = keyboard.nextInt();

to

 int temp = Integer.parseInt(keyboard.nextLine());

Answer 2

You could use keyboard.next() instead of nextLine() and as

user1770155

mentioned to compare two strings you should use .equals() and what I would do since you're comparing to an upper letter "C" is

 type = keyboard.next().toUpperCase();



if (type.equals("C"))

Answer 3

Use Scanner Class:

int temp;
java.util.Scanner s = new java.util.Scanner(System.in);
String opposite, type;
double product;

System.out.print("Please enter a temperature: ");
temp = s.nextInt();

System.out.println("Was that in Celsius or Fahrenheit?");
System.out.print("(Enter 'C' for Celsius and 'F' for Fahrenheit) ");
type = s.nextLine();

if (type.equals("C")){
do something
}
else{
do something
}

For More Input references:

Scanner

BufferedReader

String

Here is a wonderful comparison on how to compare strings in java.

Answer 4

.nextInt() does not read the end of line character "\n".

You need to put a keyboard.nextLine() after the .nextInt() and then it will work.

Answer 5

Also, use type.equals("C") instead of if (type == "C"), the later one is comparing the reference of the value.

Answer 6

For you code Change nextLine(); to next(); and it will work.

System.out.println("Was that in Celsius or Fahrenheit?");
    System.out.print("(Enter 'C' for Celsius and 'F' for Fahrenheit) ");
    type = keyboard.next();

to get an idea for you to what happened was this:

• nextLine(): Advances this scanner past the current line and returns the input that was skipped.
• next(): Finds and returns the next complete token from this scanner.

Also like the many of the answers says use equals() instead of using ==

The == checks only the references to the object are equal. .equal() compares string.

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