带有扫描仪的 Java 应用程序出错。 Next() 与 NextLine() 以及为什么我收到错误？

Question

Im having an issue with my Java application. The application is required to do the following:

(1) Prompt the user for a string that contains two strings separated by a comma. (1 pt)

Examples of strings that can be accepted: Jill, Allen Jill , Allen Jill,Allen Ex:

Enter input string: Jill, Allen

(2) Report an error if the input string does not contain a comma. Continue to prompt until a valid string is entered. Note: If the input contains a comma, then assume that the input also contains two strings. (2 pts)

Ex:

Enter input string: Jill Allen Error: No comma in string Enter input string: Jill, Allen

(3) Extract the two words from the input string and remove any spaces. Store the strings in two separate variables and output the strings. (2 pts)

Ex:

Enter input string: Jill, Allen First word: Jill Second word: Allen

(4) Using a loop, extend the program to handle multiple lines of input. Continue until the user enters q to quit. (2 pts)

Ex:

Enter input string: Jill, Allen First word: Jill Second word: Allen

Enter input string: Golden , Monkey First word: Golden Second word: Monkey

Enter input string: Washington,DC First word: Washington Second word: DC

Enter input string: q

My Code:

package parsestrings;

import java.util.Scanner;

public class ParseStrings {

    public static void main(String[] args) {
        Scanner scnr = new Scanner(System.in); // Input stream for standard input
        Scanner inSS = null;                   // Input string stream
        String lineString = "";                // Holds line of text
        String firstWord = "";                 // First name
        String secondWord = "";                  // Last name
        boolean inputDone = false;             // Flag to indicate next iteration

        // Prompt user for input
        System.out.println("Enter input string: ");

        // Grab data as long as "Exit" is not entered
        while (!inputDone) {

            // Entire line into lineString
            lineString = scnr.next();

            // Create new input string stream
            inSS = new Scanner(lineString);

            // Now process the line
            firstWord = inSS.next();

            // Output parsed values
            if (firstWord.equals("q")) {
                System.out.println("Exiting.");

                inputDone = true;

                if (firstWord.matches("[a-zA-Z]+,[a-zA-Z]+")) {
                    System.out.print("Input not two comma separated words");
                }
            } else {
                secondWord = inSS.next();

                System.out.println("First word: " + firstWord);
                System.out.println("Second word: " + secondWord);

                System.out.println();
            }
        }
        return;
    }
}

The error Im getting returned:

Exception in thread "main" java.util.NoSuchElementException
    at java.util.Scanner.throwFor(Scanner.java:862)
    at java.util.Scanner.next(Scanner.java:1371)
    at parsestrings.ParseStrings.main(ParseStrings.java:53)
Java Result: 1

Answer 1

Scanner.next doc says:

Throws: NoSuchElementException - if no more tokens are available

Reason for your exception:

lineString = scnr.next(); // this is reading just one word (token)
//...
inSS = new Scanner(lineString); // this creates a new scanner with a string consisting in just one word
firstWord = inSS.next(); // this reads the word loaded in the inSS scanner
//...
secondWord = inSS.next(); // here is the problem. There are no more words left in this scanner (tokens) and throws the exception.

So It would work if you change this line:

lineString = scnr.next();

with this:

lineString = scnr.nextLine();

Scanner.nextLine will read the entire line as you are expecting. Anyway the user can input one word. So it would be nice to validate the input before proceding. You can do it like this:

lineString = scnr.nextLine();

if(lineString == null || lineString.length() == 0 || lineString.split("\\s+").length < 2){
    System.out.println("2 words required. Try again:");
    continue;
}

Here I'm using a regular expression to validate that there is more than one word in the input.