Scanner Is Skipping nextLine() After Using Next() or nextFoo()?

Question

I am using the Scanner methods nextInt() and nextLine() for reading input.

It looks like this:

System.out.println("Enter numerical value");    
int option;
option = input.nextInt(); // Read numerical value from input
System.out.println("Enter 1st string"); 
String string1 = input.nextLine(); // Read 1st string (this is skipped)
System.out.println("Enter 2nd string");
String string2 = input.nextLine(); // Read 2nd string (this appears right after reading numerical value)

The problem is that after entering the numerical value, the first input.nextLine() is skipped and the second input.nextLine() is executed, so that my output looks like this:

Enter numerical value
3   // This is my input
Enter 1st string    // The program is supposed to stop here and wait for my input, but is skipped
Enter 2nd string    // ...and this line is executed and waits for my input

I tested my application and it looks like the problem lies in using input.nextInt(). If I delete it, then both string1 = input.nextLine() and string2 = input.nextLine() are executed as I want them to be.

Accepted Answer

That's because the Scanner.nextInt method does not read the newline character in your input created by hitting "Enter," and so the call to Scanner.nextLine returns after reading that newline.

You will encounter the similar behaviour when you use Scanner.nextLine after Scanner.next() or any Scanner.nextFoo method (except nextLine itself).

Workaround:

Either put a Scanner.nextLine call after each Scanner.nextInt or Scanner.nextFoo to consume rest of that line including newline
```
int option = input.nextInt();
input.nextLine();  // Consume newline left-over
String str1 = input.nextLine();
```
Or, even better, read the input through Scanner.nextLine and convert your input to the proper format you need. For example, you may convert to an integer using Integer.parseInt(String) method.
```
int option = 0;
try {
    option = Integer.parseInt(input.nextLine());
} catch (NumberFormatException e) {
    e.printStackTrace();
}
String str1 = input.nextLine();
```

Answer 2

The problem is with the input.nextInt() method - it only reads the int value. So when you continue reading with input.nextLine() you receive the "\n" Enter key. So to skip this you have to add the input.nextLine(). Hope this should be clear now.

Try it like that:

System.out.print("Insert a number: ");
int number = input.nextInt();
input.nextLine(); // This line you have to add (It consumes the \n character)
System.out.print("Text1: ");
String text1 = input.nextLine();
System.out.print("Text2: ");
String text2 = input.nextLine();

Answer 3

It's because when you enter a number then press Enter, input.nextInt() consumes only the number, not the "end of line". When input.nextLine() executes, it consumes the "end of line" still in the buffer from the first input.

Instead, use input.nextLine() immediately after input.nextInt()

Answer 4

There seem to be many questions about this issue with java.util.Scanner. I think a more readable/idiomatic solution would be to call scanner.skip("[\r\n]+") to drop any newline characters after calling nextInt().

EDIT: as @PatrickParker noted below, this will cause an infinite loop if user inputs any whitespace after the number. See their answer for a better pattern to use with skip:

Answer 5

It does that because input.nextInt(); doesn't capture the newline. you could do like the others proposed by adding an input.nextLine(); underneath.
Alternatively you can do it C# style and parse a nextLine to an integer like so:

int number = Integer.parseInt(input.nextLine());

Doing this works just as well, and it saves you a line of code.

Answer 6

TL;DR
Use scanner.skip("\\R?") before scanner.nextLine() if you are not sure if previous instruction was scanner.next() or scanner.next*TypeName*().

skip uses regex where

\R represents line separators
? will make \R line separator sequence optional - which will prevent skip method from

(a) waiting for matching sequence - in case of reaching end of still opened source of data like System.in
(b) throwing java.util.NoSuchElementException in case of terminated/closed source of data, or when actual data doesn't match what we want to skip.

Must Read

Things you need to know:

text which represents few lines also contains non-printable characters between lines (we call them line separators) like
carriage return (CR - in String literals represented as "\r")
line feed (LF - in String literals represented as "\n")
when you are reading data from the console, it allows the user to type his response and when he is done he needs to somehow confirm that fact. To do so, the user is required to press "enter"/"return" key on the keyboard.

What is important is that this key beside ensuring placing user data to standard input (represented by System.in which is read by Scanner) also sends OS dependant line separators (like for Windows \r\n) after it.

So when you are asking the user for value like age, and user types 42 and presses enter, standard input will contain "42\r\n".

Scanner Is Skipping nextLine() After Using Next() or nextFoo()?

23 Answers

Must Read

Things you need to know:

Sophia Al-Mansoor