我想从.txt文件中读取有关MMORPG字符的数据,然后过滤体验(最低经验)。但是我得到了这个例外,我知道它的含义,但我真的不明白我做错了什么。
我不擅长java,我实际上是初学者。有人可以向我解释一下吗?可能我做的事非常愚蠢。
这是我的代码:
排名(汉字)
import java.util.Scanner;
public class Karakter {
private String name;
private int experience;
private int maxHealthPoints;
private int healthPoints;
private int maxGreed;
private int greed;
public Karakter(String nm, int exp, int mHP, int hp, int mG, int g)
{
name = nm;
experience = exp;
maxHealthPoints = mHP;
healthPoints = hp;
maxGreed = mG;
greed = g;
}
public String toString()
{
String s = "";
s += getName();
s += getExperience();
s += getMaxHealthPoints();
s += getHealthPoints();
s += getMaxGreed();
s += getGreed();
return s;
}
public static Karakter read(Scanner sc)
{
String name = sc.next();
int experience = sc.nextInt();
int maxHealthPoints = sc.nextInt();
int healthPoints = sc.nextInt();
int maxGreed = sc.nextInt();
int greed = sc.nextInt();
return new Karakter(name, experience, maxHealthPoints, healthPoints, maxGreed, greed);
}
public boolean hasExperience(int min)
{
return experience >= min;
}
// returns true if Krakters have the same name
public boolean equals(Object other)
{
if(!(other instanceof Karakter))
{
return false;
}
else
{
Karakter that = (Karakter) other;
return that.name == this.name;
}
}
public String getName()
{
return name;
}
public int getExperience()
{
return experience;
}
public int getMaxHealthPoints()
{
return maxHealthPoints;
}
public int getHealthPoints()
{
return healthPoints;
}
public int getMaxGreed()
{
return maxGreed;
}
public int getGreed()
{
return greed;
}
}
字符(字符)
import java.io.FileReader;
import java.io.IOException;
import java.util.ArrayList;
import java.util.Scanner;
public class Karakters {
private ArrayList<Karakter> kars = new ArrayList<Karakter>();
public void voegToe(Karakter kar)
{
if(!kars.contains(kar))
{
kars.add(kar);
}
}
// returns an arraylist of characters with exp >= minexp
public ArrayList<Karakter> karaktersVanaf(int minExperience)
{
Karakter kar = null;
for(int i = 0; i < kars.size(); i++)
{
if(kar.hasExperience(minExperience))
kars.add(kar);
}
return kars;
}
public static Karakters read(String infile)
{
Karakters k = new Karakters();
try
{
FileReader fr = new FileReader(infile);
Scanner sc = new Scanner(fr);
int aantal = sc.nextInt();
for(int i = 0; i < aantal ; i++)
{
Karakter kar = Karakter.read(sc);
k.kars.add(kar);
}
fr.close();
}
catch(IOException iox)
{
System.out.println(iox);
return null;
}
return k;
}
}
KarakterZoeker(主要方法):
import java.util.ArrayList;
import java.util.Scanner;
public class KarakterZoeker {
public static void main(String[] args)
{
Scanner sc = new Scanner(System.in);
Karakters kars = Karakters.read("infile.txt");
System.out.println("Welke experience wilt u minimaal?");
int minExp = sc.nextInt();
ArrayList<Karakter> s = kars.karaktersVanaf(minExp);
System.out.println(s.toString());
}
}
这是错误:
Exception in thread "main" java.util.InputMismatchException
at java.util.Scanner.throwFor(Unknown Source)
at java.util.Scanner.next(Unknown Source)
at java.util.Scanner.nextInt(Unknown Source)
at java.util.Scanner.nextInt(Unknown Source)
at Karakter.read(Karakter.java:38)
at Karakters.read(Karakters.java:41)
at KarakterZoeker.main(KarakterZoeker.java:14)
这些线:
Karakter kar = Karakter.read(sc); (in Karakters)
int experience = sc.nextInt(); (in Karakter)
Karakters kars = Karakters.read("infile.txt"); (in KarakterZoeker)
这是txt文件:
2
name = Dursley
experience = 15
maxHealthPoints = 20
healthPoints = 10
maxGreed = 3
greed = 1
name = Aragorn
experience = 45
maxHealthPoints = 40
healthPoints = 30
maxGreed = 20
greed = 10
问题是在这一行:
int experience = sc.nextInt();
您正在尝试读取“int”,但扫描仪读取的是字符串"experience"
从.txt文件中的这一行:
experience = 15
所以你必须阅读两个字符串(第一个是“经验”,第二个是“=”)然后你将能够读取一个int(15),依此类推。
第一个没有失败,因为name
被分配了一个字符串,但如果你记录你所读的内容,你就得到了:“name”而不是真名。
另一种方法是让ResourceBundle
班为你做读。甚至是Properties
班。
使用ResourceBundle,您可以执行以下操作:
ResourceBundle bundle = ResourceBundle.getBundle("Karakters");
String exp = bundle.getString("experience");
int experience = Integer.parseInt( exp );
你必须将其命名为:Karakters.properties
编辑
我已经添加了一个关于如何使用ResouceBundle
执行此操作的示例
扫描程序抛出InputMismatchException
以指示检索到的令牌与预期类型的模式不匹配,或者令牌超出预期类型的范围。
所以,我的猜测是,当下一个标记实际上不是nextInt
时,你正在调用Integer
。
编辑
阅读该文件时需要小心,因为您的标签被视为标记。例如,当你有:
maxHealthPoints = 20 ,
你有三个令牌:“maxHealthPoints”,“=”和“20”。问题是你忽略了标签(在“=”之前)和等号。我的建议?逐行读取文件并使用Scanner
解析每一行。或者,从文件中删除标签和等号,只留下要提取的值。
两次调用next()
解决了这个问题。之后我在卡拉克特斯的NullPointerException
得到了一个karaktersVanaf(minExp)
。为了解决这个问题,我在代码中添加了这些行:
public ArrayList<Karakter> karaktersVanaf(int minExperience) throws FileNotFoundException
{
Scanner sc = new Scanner(new FileReader("infile.txt")); // --------line 1
Karakter kar = Karakter.read(sc); // ------------ line 2
for(int i = 0; i < kars.size(); i++)
{
if(kar.hasExperience(minExperience))
kars.add(kar);
}
return kars;
}
但现在我再次看到InputMissMatschException
!我不知道我做错了什么!我可能正在做一些完全错误的事情,所以要准备好笑。
顺便说一句,我还没有15点经验值,这就是我现在不能为你投票的原因。我会尽快这样做。
每次读者执行nextInt()或nextString()时,它都不会读取下一行,它会读取由空格(空格或换行符)分隔的下一个字符。解决问题的最简单方法是删除“name =”,“experience =”,然后将唯一数据放在每一行上。
所以你的文件看起来像:
2
Dursley
15
20
10
3
1
Aragorn
45
40
30
20
10
然后你的代码将按原样运行。
我编写了一种使用ResourceBundle读取的方法。
我把它留在这里,我希望你发现它有用:
import java.util.*;
public class Ktr {
public static void main( String [] args ) {
ResourceBundle b = ResourceBundle.getBundle("k");
String names = b.getString("names");
List<K> l = new ArrayList<K>();
for( String name : names.split(",")){
l.add( K.read( name, b ));
}
System.out.println( l );
}
}
class K {
private String name;
private int maxHealthPoints;
private int maxGreed;
// etc..
public static K read( String name, ResourceBundle b ) {
K k = new K();
k.name = name;
k.maxHealthPoints =
Integer.parseInt(b.getString( name + ".maxHealthPoints") );
k.maxGreed =
Integer.parseInt(b.getString( name + ".maxGreed") );
// you get the idea
return k;
}
public String toString() {
return String.format("%s{maxHealthPoints=%d, maxGreed=%d}",
name, maxHealthPoints, maxGreed);
}
}
可.properties
names=Dursley,Aragorn
Dursley.maxHealthPoints=20
Dursley.maxGreed=3
Aragorn.maxHealthPoints=40
Aragorn.maxGreed=20
输出:
java Ktr
[Dursley{maxHealthPoints=20, maxGreed=3}, Aragorn{maxHealthPoints=40, maxGreed=20}]