我收到输入不匹配错误,但我不确定原因。它正在一台不同的计算机上工作,但它似乎不能在我的笔记本电脑上工作。首先我认为它可能是我的txt文件,但这不是原因。
import java.io.*;
import java.util.*;
public class FindGrade {
public static final int NUM_SCORE_TYPES = 5;
public static void main(String[] args) {
Scanner scan = null;
int[] quizArray = null;
int[] labArray = null;
int[] attendance = null;
int[] midterms = null;
int quizgrade =0;
int labgrade=0;
int attendance_1=0;
int midterms_1 =0;
String name;
try {
scan = new Scanner(new File("input.txt"));
} catch (FileNotFoundException e) {
e.printStackTrace();
return;
}
// each iteration is for single exam type (ie: Quizzes is the 1st one)
for (int i = 0; i < NUM_SCORE_TYPES; i++) {
name = scan.next();
int numScores = scan.nextInt();
int maxGrade = scan.nextInt();
if (name.equals("Quizzes")) {
quizArray = new int[numScores];
readScores(quizArray, numScores, scan);
}
else if (name.equals("Labs")) {
labArray = new int[numScores];
readScores(labArray, numScores, scan);
}
else if (name.equals("Lab_attendance")) {
attendance = new int[numScores];
readScores(attendance, numScores, scan);
}
else if (name.equals("Midterms")) {
midterms = new int[numScores];
readScores(midterms, numScores, scan);
}
}
}
public static void readScores(int[] scoreArray, int numScores, Scanner scan) {
for (int i = 0; i < numScores; i++) {
scoreArray[i] = scan.nextInt();
}
}
public static void average(double [] scoreArray, int numScores){
double sum=0;
for(int i=0; i< scoreArray.length; i++){
sum += scoreArray[i];
}
double average = sum/numScores;
System.out.println(sum + " " + average);
}
}
输出:
Exception in thread "main" java.util.InputMismatchException
at java.util.Scanner.throwFor(Scanner.java:840)
at java.util.Scanner.next(Scanner.java:1461)
at java.util.Scanner.nextInt(Scanner.java:2091)
at java.util.Scanner.nextInt(Scanner.java:2050)
at FindGrade.main(FindGrade.java:33)
输入文件:
Quizzes 8 10
5 8 9 10 4 0 10 7
Labs 6 100
95 90 100 87 63 92
Lab_attendance 16 1
1 1 1 0 1 1 1 1 0 1 1 1 1 0 1 1
Midterms 2 100
87 94
Final 0 100
答案 0 :(得分:0)
您的代码与您发布的数据一起正常运行。 如果变量“name”将被赋予与您期望的5个名称不同的值,它将给出此错误,例如“Quizze”而不是“Quizzes”。 在这种情况下,您的代码逻辑会被破坏,您将遇到此错误。这是一个可能的根本原因。