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.
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.
A Scanner breaks its input into tokens using a delimiter pattern, which by default matches whitespace. next() : Finds and returns the next complete token from this scanner. nextLine() : Advances this scanner past the current line and returns the input that was skipped.
In Scanner class if we call nextLine() method after any one of the seven nextXXX() method then the nextLine() doesn't not read values from console and cursor will not come into console it will skip that step.
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();
If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With