Question

我的代码应该从.txt文件中读取输入，然后执行不同的操作来对其进行排序。第一个int是长度。我的主要问题是我无法弄清楚为什么我一直收到不匹配错误，或者如果我输入'垃圾'语句，程序永远不会完成。我先发布.txt，然后发布程序。

15
Smith, John
26
Baker
Jones, Susan
15
Student
Mouse, Mickey
31
Theme park employee
Mouse, Mighty
48
Cartoon super hero
Anderson, William
35
Computer Programmer
Parker, Cindy
18
Author
McCain, John
20
Student
Armstrong, Michelle
17
Student
Thompson, Anne
29
Doctor
Li, Steve
15
Student
James, Tanya
20
Student
Moore, James
32
Teacher
Andrews, Julie
75
Actress
Obama, Michelle
46
Lawyer
Michaels, Todd
51
Student

//不要忘记在最后复制空行。

程序从这里开始。

import java.util.Scanner;
import java.io.*;
public class SortAndDisplayCustomerData
{
public int length; //The length of the names, ages, and occupations arrays
public String[] names;
public int[] ages;
public String[] occupations;
public int count; //The length of the studentNames and studentAges arrays
public String[] studentNames;
public int[] studentAges;
public int i, minPos, Temp2, y, minVal;
public String Temp, Temp3, temp2, minVal2;

public void getDataFromFile()
{
Scanner keyboard = new Scanner(System.in);
    Scanner inputStream = null;
System.out.println("wtf");
    try
    {
        inputStream = new Scanner(new FileInputStream("NameAgeOcc.txt"));
    }

    catch(FileNotFoundException error)
    {
        System.out.println("Unable to open input file.");
        System.exit(0);
    }
    System.out.println("wtf2");
    length=inputStream.nextInt();
    //String junk = keyboard.nextLine();
    System.out.println("wtf3");
    names = new String[length];
    ages = new int[length];
    occupations = new String[length];
    for(i=0;i<length;i++)
    {
        names[i]=inputStream.nextLine();
        ages[i]=inputStream.nextInt();
        occupations[i]=inputStream.nextLine();
    }
    inputStream.close();
} 

public void displayAllFileData()
{
System.out.println("wtf3");
System.out.printf("%-25s%-8s%24s%n","Names","  Ages","  Occupations");
    for(i=0;i<length;i++)
    {
        System.out.printf("%-25s%6d%-24s%n",names[i],ages[i],occupations[i]);
    }
}
public void sortAllDataByAge()
{
    for(i=0;i<length;i++)
    {
        minVal=ages[i];
        minPos=i;
        for(y=i+1;y<length;y++)
        {
            if(ages[y]<minVal)
            {
                minVal=ages[y];
                minPos=y;
            }
        }
        Temp2 = ages[minPos];
        ages[minPos] = ages[i];
        ages[i] = Temp2;
        Temp = names[minPos];
        names[minPos] = names[i];
        names[i] = Temp;
        Temp3 = occupations[minPos];
        occupations[minPos] = occupations[i];
        occupations[i] = Temp3;
    }
}
public void extractStudentData()
{
    count=0;
    for (i=0;i<length;i++)
    {
        if(occupations[i].equalsIgnoreCase("student"))
        count++;
    }
    int j=0;
    studentAges = new int[count];
    studentNames = new String[count];
    for (i=0;i<length;i++)
    {
        if(occupations[i].equalsIgnoreCase("student"))
        {
            studentAges[j]=ages[i];
            studentNames[j]=names[i];
            j++;
        }
        }
}
public void displayStudentData()
{ 
System.out.printf("%n%-25s%-8s%n","Names","  Ages");

    for (i=0;i<count;i++)
    {
        System.out.printf("%-25s%6d%n",studentNames[i],studentAges[i]);
    }

}
public void sortStudentDataAlpha()
{ 
    for(i=0;i<count;i++)
    {
        minVal2=studentNames[i];
        minPos=i;
        for(y=i+1;y<count;y++)
        {
            if(studentNames[y].compareToIgnoreCase(minVal2)<0)
            {
                minVal2=studentNames[y];
                minPos=y;
            }
        }
        Temp = studentNames[minPos];
        studentNames[minPos] = studentNames[i];
        studentNames[i] = Temp;
        Temp2 = studentAges[minPos];
        studentAges[minPos] = studentAges[i];
        studentAges[i] = Temp2;
    }
}

}

Answer 1

在nextInt（）之后你应该吃掉换行符，否则你会期待一个字符串（对于文本文件中的名字）并且你会得到一个空字符串（换行符））。然后你再次有一个nextInt（），但文本文件中的指针将在名称（String）上。

例如，您的程序遵循以下路径：

32 
Teacher 
Andrews, Julie 
75 
Actress 
Obama, Michelle

nextInt（） - ＆gt; 32 nextLine（） - ＆gt; nextLine（） - ＆gt; Teacher
nextInt（） - ＆gt; Andrews, [类型不匹配]

java不匹配或永不结束的程序

1 个答案: