Taking string as input and converting to time/date in java

Question

When I take the string input using next(), I don't get any exception


String a=sc.next()
SimpleDateFormat b=new SimpleDateFormat("dd/MM/yyyy");
b.parse(a);
System.out.println(b.format(b.parse(a)));

But when I take the input using nextLine() method of Scanner class, I get an exception

String a=sc.nextLine()
SimpleDateFormat b=new SimpleDateFormat("dd/MM/yyyy");
b.parse(a);
System.out.println(b.format(b.parse(a)));

How to fix this, I get the same exception when using sc.nextLine() and date format as "hh:mm a"

What exception are you getting? What does the scanner actually read in? is it user input or are you reading data from a file? — OH GOD SPIDERS, Feb 08 '21 at 17:08
It is user input...java.text.ParseException: Unparseable date: ""...This is the exception I'm getting.Actually the question says " The first line of each test case contains a single time P " and the time is of the form "hh:mm a" (eg 10:43 AM)...Can't figure out how to take this input. — nandamuri pavan, Feb 08 '21 at 17:31
O please, stop using `SimpleDateFormat`. That class, just like `Date` and `Calendar`, is outdated and [troublesome](https://stackoverflow.com/questions/1969442/whats-wrong-with-java-date-time-api). Use classes from the `java.time` package instead. — MC Emperor, Feb 08 '21 at 18:07
Welcome to Stack Overflow. For better reception of your next questions you will want to read [How do I ask a good question?](https://stackoverflow.com/help/how-to-ask) and [How to create a Minimal, Reproducible Example](https://stackoverflow.com/help/minimal-reproducible-example) before posting. — Ole V.V., Feb 08 '21 at 19:44

score 1 · Answer 1 · 2021-02-09T09:39:18.110

This problem maybe related to how the Scanner works.
See documentation of next:

Finds and returns the next complete token from this scanner. A complete token is preceded and followed by input that matches the delimiter pattern.

and of nextLine:

Advances this scanner past the current line and returns the input that was skipped.

The default delimiter includes the newline character (\n).

I assume the input has a leading newline (empty line) that is being removed by next but not by nextLine.

Test:

new Scanner("\ntest\n").next() returns "test"
new Scanner("\ntest\n").nextLine() returns "" (empty string)

This will happen, for example, if the previous Scanner input was read with nextInt() (or similar) - this does not consume any newline after the number:

var sc = new Scanner("123\n456\n");
var i = sc.nextInt();  // will be 123
var str = sc.nextLine();  // will be the empty string "" (between 123 and \n)
str = sc.nextLine();  // now it is "test"

_{String used as input for the Scanner since it is easier to describe, same will happen with user input where a newline will be read if user types the return key (usually to end input)}

If this is the case, solution is to call nextLine() after reading a number _{(or even read the number with nextLine() and then parsing it)}.

For more details, please check this question: Scanner is skipping nextLine() after using next() or nextFoo()?

_{this answer is based on the comment that the Exception mentions the empty string "" as unparseable}
also consider vsfDawg's answer - next() will stop at whitespace character

BTW `"dd/MM/yyyy"` will not parse time – Feb 08 '21 at 18:16 — , Feb 08 '21 at 18:16

score 0 · Answer 2 · answered Feb 08 '21 at 18:00

If your input looks like this:

10/12/2020 11:59:59 am

Then the first invocation of the Scanner next() method would return "10/12/2020" while the first invocation of nextLine() would return "10/12/2020 11:59 am". The former matches your DateFormat while the latter does not. That is may be why next() is behaving but you get an Exception with nextLine().

Taking string as input and converting to time/date in java

2 Answers2