我在循环中的Java程序中有一个运行的时间错误

Question

代码编译，但是当我执行它时，错误说

" java.util.nosuchelementException"，并突出显示湿度= input.next（）;

样本输出如下：

K-Nearest Neighbor Prediction Program:
The values you entered are: sunny cool high true
Comparring the values:  sunny  hot   high false dist = 2 and play = no
Comparring the values:  sunny  hot  high true  dist = 1 and play = no
Comparring the values:  overcast hot  high false dist = 3 and play = yes
Comparring the values:  rainy mild  high false dist = 3 and play = yes
Comparring the values:  rainy cool  normal false dist = 3 and play = yes

这是我的代码：

import java.util.Scanner;
public class lab2
{
    public static void main (String [] args)
    {
        Scanner input = new Scanner ("data.txt");
        System.out.println("The values you entered are : ");
        String queryOutlook = "sunny";
        String queryHumidity = "high";
        String queryTemp = "cool";
        String queryVerd = "true";
        String outlook = null;
        String humidity = null;
        String temp = null;
        String verd = null;
        String play= null;
        int distance = 0;
        while(input.hasNext())
        {
            outlook = input.next();
            ***humidity = input.next();***
            temp = input.next();
            verd = input.next();
            if (!outlook.equalsIgnoreCase(queryOutlook) )
            distance++;
            if (!humidity.equalsIgnoreCase(queryHumidity) )
            distance++;
            if (!temp.equalsIgnoreCase(queryTemp) )
            distance++;
            if (!verd.equalsIgnoreCase(queryVerd) )
            distance++;

            System.out.println(outlook + humidity + temp + verd + "t" + play + distance);

再次，该程序与自上课时期以前从未听说过的K-Neartible邻居预测有关。任何帮助，将不胜感激！！！

Answer 1

    while(input.hasNext()) {
        outlook = input.next();
        humidity = input.next();
        temp = input.next();
        verd = input.next();

您正在检查输入是否具有下一个元素，如果确实有一个元素，则将获得四个下一个元素。您应该在获取每一个之前检查。

Answer 2

scanner（）的构造函数以java.io.file作为输入。您将其传递给字符串。因此，您的扫描仪仅扫描匹配的字符串" data.txt"，它实际上从未打开该文件并扫描它。

您需要做类似的事情才能开始：

File f = new File("data.txt");
Scanner input = new Scanner(f);
... // and so on