Question

我编写了这段代码，用于读取具有整数值的文件。如果整数值是> = 0且＆lt; = 100，我需要给出等级的平均值。如果有超出指定范围0-100的任何值，那么我需要计算不正确的整数等级，告知用户不正确的等级，并告知有多少不正确的等级。我尝试了代码，但我一直收到错误代码：

Exception in thread "main" java.util.NoSuchElementException
at java.util.Scanner.throwFor(Unknown Source)
at java.util.Scanner.next(Unknown Source)
at Project9.main(Project9.java:26)

代码示例：

public static void main(String[] args) throws IOException{
    String file;
    int readInts;

    Scanner k = new Scanner(System.in);

    System.out.println("Enter filename: ");
    file = k.nextLine();
    int counterWrong = 0;
    int counterRight = 0;
    int sum = 0;
    double average = 1.0 * sum/counterRight;

    File fileReader = new File(file);

    if (fileReader.exists()) {
        Scanner input = new Scanner(fileReader);
        while (input.hasNext()) {
            readInts = input.nextInt();
            System.out.println(readInts);
            String a = input.next();
            int a2 = Integer.parseInt(a);

        if (a2 <= 100 && a2 >= 0){
            counterRight++;            
            sum = sum + a2; 
            System.out.println("Score " + a2 + " was counted.");

        } else {
            counterWrong++;
            System.out.println("The test grade " + a2 + " was not scored as it was out of the range of valid scores.");
            System.out.println("There were " + counterWrong + " invalid scores that were not counted.");
        }
        }
        if (counterRight > 0){
            System.out.println("The average of the correct grades on file is " + average + ".");
        }
    } else {
        System.out.println("The file " + file + " does not exist. The program will now close.");
    }


}

}

Answer 1

您正在进行一次检查hasNext，但之后您使用nextInt()和next()从扫描仪中读取了两次。

Answer 2

使用hasNextInt（）而不是hasNext（）。

hasNext（）只表示存在另一个标记，不一定是您在编写nextInt（）时假设的另一个整数。

以下是hasNext()和hasNextInt()

的文档

您还想在此行之前进行检查：

String a = input.next();

Answer 3

我看到您的代码可能存在两个问题。

file = k.nextLine（）; //根据文件的设置方式，k.nextLine（）或k.next（）或者k.nextInt（）可能会有用。
while（input.hasNext（））{ readInts = input.nextInt（）; // input.hasNext（）假设扫描器正在读取的下一个值具有一个字符串值，这将使readInts = input.nextInt（）;没有解析（或其他方法）就不可能使用。

我认为尝试这个练习很有趣（不想为你毁掉它）。查看我的代码，希望您能了解我正在讨论的一些概念。

注意：我的程序读取整数值，如95 185 23 13 90 93 37 125 172 99 54 148 53 36 181 127 85 122 195 45 79 14 19 88 34 73 92 97 200 167 126 48 109 38.哪些使用hasNext（）＆amp; next（）获取列出的每个标记。因此，使用nextLine（）对于给定的输入不会有用。

包cs1410;

import java.io.File;
import java.io.IOException;
import java.util.Scanner;

import javax.swing.JFileChooser;

public class Grader {

    public static void main(String[] args) throws IOException {
        int count = 0;
        int sum = 0;
        double ave = 0;
        int incorrectCount = 0;
        String correctGrades = "";
        String incorrectGrades = "";

        // Read file input
        JFileChooser chooser = new JFileChooser();
        if (JFileChooser.APPROVE_OPTION != chooser.showOpenDialog(null)) {
            return;
        }
        File file = chooser.getSelectedFile();

        // Scan chosen document
        Scanner s = new Scanner(file);


        // While the document has an Int
        while (s.hasNextInt()) {
            // Convert our inputs into an int
            int grade = Integer.parseInt(s.next());

            if (grade >= 0 && grade <= 100) {
                // adds sum
                sum += grade;
                // increments correct count
                count++;
                // displays valid grades
                correctGrades += Integer.toString(grade) + "\n";
            } else {
                // increments incorrect count
                incorrectCount++;
                // displays invalid grades
                incorrectGrades += Integer.toString(grade) + "\n";
            }
        }
        // Created average variable 
        ave = sum / count;

        // bada bing bada boom
        System.out.println("The number of correct grades were " + correctGrades);
        System.out.println("The average score on this test was " + ave + "\n");
        System.out.println("The number of incorrect grades were " + incorrectCount + "\n");
        System.out.println("The incorrect values for the grades were " + "\n" + incorrectGrades);

    }

}

Java循环混乱

3 个答案: