在java中打印一个列表

Question

我需要从文件中读取一些数组并打印它们。我的第一个课程处理菜单驱动的程序，用户输入一个数字来告诉程序连接到文件，打印文件中的名称，整数，字符或双打。我坚持的第一件事是连接到文件。这是我的不完整的类从文件读取：

import java.util.*; 
public class Prog5Methods{ 
public Prog5Methods(){ 
} 
    public void ReadFromFile(Scanner input, String [] names, int [] numbers, char [] letters, double [] num2){ 
     System.out.println("\nReading from a file...\n"); 
     System.out.println("\nDONE\n"); 
     int r = 0; 
      while(input.hasNext()){ 
       names[r] = input.next(); 
       numbers[r] = input.nextInt(); 
       letters[r] = input.next().charAt(0); 
       num2[r] = input.nextDouble(); 
       r++; 
     } 
} // end of readFromFile 




} 

这是我从读文件包含：

Lee Keith Austin Kacie Jason Sherri  Jordan  Corey Reginald Brian Taray 
Christopher Randy Henry Jeremy Robert Joshua Robert Eileen 
Cassandra Albert Russell Ethan Cameron Tyler Alex Kentrell rederic 
10 20 100 80 25 35 15 10 45 55 200 300 110 120 111 7 27 97 17 37 
21 91 81 71 16 23 33 45 
A b c w e r t q I u y b G J K S A p o m b v x K F s q w 
11.5 29.9 100 200 115.1 33.3 44.4 99.9 100.75 12.2 13.1 20.3 55.5 77.7 
12.1 7.1 8.2 9.9 100.1 22.2 66.6 9.9 1.25  3.75 19.9 3.321 45.54 88.8 

的名字是数组名[]，该整数是在阵列数字[]等。我需要从这些数组中打印每个变量。

Answer 1

使用List<T>代替阵列。从基础流使用扫描仪

public static void ReadFromFile(Scanner input, 
     ArrayList<String> names, 
     ArrayList<Integer> numbers, 
     ArrayList<Character> letters, 
     ArrayList<Double> num2) 
{ 

    while(input.hasNext()) 
    { 
    String val=input.next(); 
    Object no=parseInt(val); 
    if(no!=null) //Is integer? 
     { 
      numbers.add((Integer)no); 
     } 
    else 
    { 
     no=parseDouble(val); 
     if(no!=null) // Is double? 
     { 
      num2.add((Double)no); 
      } 
     else 
     { 
      no=parseChar(val); 
      if(no!=null) //Is Char? 
      { 
      letters.add((Character)no); 
      } 
      else 
      { 
       names.add(val); // String 
      } 
     } 
     } 
    } 
} 

方法解析字符串

读取值。

public static Integer parseInt(String str) 
    { 
     Integer retVal=-1; 
     try 
     { 
      retVal=Integer.parseInt(str); 
     }catch(Exception ex) { return null;} 
     return retVal; 
    } 
    public static Double parseDouble(String str) 
    { 
     double retVal=-1; 
     try 
     { 
      retVal=Double.parseDouble(str); 
     }catch(Exception ex) { return null;} 
     return retVal; 
    } 
    public static Character parseChar(String str) 
    { 
     Character retVal=null; 

     if(str.length()==1) 
      retVal=str.charAt(0); 
     return retVal; 
    } 

测试你的代码

public static void main(String[] args) throws Exception 
     { 
     ....... 
     ArrayList<String> names=new ArrayList<String>(); 
     ArrayList<Integer> numbers=new ArrayList<Integer>(); 
     ArrayList<Double> num2=new ArrayList<Double>(); 
     ArrayList<Character> letters=new ArrayList<Character>(); 

     ReadFromFile(input,names,numbers,letters,num2); 

     System.out.println(names); 
     System.out.println(numbers); 
     System.out.println(letters); 
     System.out.println(num2); 
     } 

Answer 2

对于从文本文件中读取数据，FileReader是您的最佳选择;

BufferedReader b = new BufferedReader(new FileReader("filename.txt")); 
String s = ""; 
while((s = b.readLine()) != null) { 
    System.out.println(s); 
} 

会读取文件中的每一行，并一次将其打印到标准输出一行。

import java.io.*; 
public class Prog5 { 
    public static String names[]; 
    public static int integers[]; 
    public static char letters[]; 
    public static float decimals[]; 
    public void readFileContents() { 
     File f = new File("yourfilename.txt"); 
     byte b = new byte[f.length()]; 
     FileInputStream in = new FileInputStream(f); 
     f.read(b); 
     String wholeFile = new String(b); 
     String dataArray[] = wholeFile.split(" "); 
     for(int i = 0; i < dataArray.length; i++) { 
     String element = dataArray[i]; 
     //in here you need to figure out what type element is 
     //or you could just count a certain number or each type if you know in advance 
     //then you need to parse it with eg Integer.parseInt(element); for the integers 
     //and put it into the static arrays 
     } 
    } 
} 

Answer 3

您需要周围的文件名 '新文件'，创建一个扫描仪，

p5m.readFromFile (new Scanner (new File("user.data")), 

要使用它，你需要java.io：

import java.io.*; 

如果我坚持您的要求，使用数组，和扫描仪，我可以做这样的事情：

import java.util.*; 
import java.io.*; 

public class Prog5Methods 
{ 
    public void readFromFile (Scanner input, String [] names, int [] numbers, char [] letters, double [] num2) 
    { 
     System.out.println("\nReading from a file...\n"); 
     String [][] elems = new String [5][]; 
     for (int i = 0; i < 4; ++i) 
     { 
      elems[i] = input.nextLine().split ("[ \t]+"); 
     } 

     for (int typ = 0; typ < 4; ++typ) 
     { 
      int i = 0; 
      for (String s: elems[typ]) 
      { 
       switch (typ) 
       { 
        case 0: names [i++] = s; break; 
        case 1: numbers[i++] = Integer.parseInt (s); break; 
        case 2: letters[i++] = s.charAt (0); break; 
        case 3: num2[i++] = Double.parseDouble (s); break; 
       } 
       System.out.println (i + " " + typ + " " + s); 
      } 
     } 
     System.out.println("\nDONE\n"); 
    } 

    public static void main (String args[]) throws FileNotFoundException 
    { 
     Prog5Methods p5m = new Prog5Methods(); 
     p5m.readFromFile (new Scanner (new File("user.data")), 
      new String [28], 
      new int [28], 
      new char [28], 
      new double [28]); 
    } 
} 

问题1：我需要知道，每行有28个元素，并且我将它们打印上的苍蝇，在这里。它们被卡在匿名数组中，从未使用过，但这只适用于简短的演示。我可以宣布的阵列，稍后打印：

String [] names = new String [28]; 
    int [] numbers = new int [28]; 
    char [] letters = new char [28]; 
    double [] num2 = new double [28]; 

    Prog5Methods p5m = new Prog5Methods(); 
    p5m.readFromFile (new Scanner (new File("user.data")), 
     names, numbers, letters, num2); 

我仍被绑28个元素。我可以延迟数组的初始化，直到通过读取文件知道有多少元素。

public String [][] readFromFile (Scanner input) 
{ 
    System.out.println("\nReading from a file...\n"); 
    String [][] elems = new String [5][]; 
    for (int i = 0; i < 4; ++i) 
    { 
     elems[i] = input.nextLine().split ("[ \t]+"); 
    } 
    return elems; 
} 

public static void main (String args[]) throws FileNotFoundException 
{ 
    Prog5Methods p5m = new Prog5Methods(); 
    String [][] elems = p5m.readFromFile (new Scanner (new File("user.data"))); 

    int size = elems[0].length; 

    String [] names = new String [size]; 
    int [] numbers = new int [size]; 
    char [] letters = new char [size]; 
    double [] num2 = new double [size]; 

    for (int typ = 0; typ < 4; ++typ) 
    { 
     int i = 0; 
     for (String s: elems[typ]) 
     { 
      switch (typ) 
      { 
       case 0: names [i++] = s; break; 
       case 1: numbers[i++] = Integer.parseInt (s); break; 
       case 2: letters[i++] = s.charAt (0); break; 
       case 3: num2[i++] = Double.parseDouble (s); break; 
      } 
     } 
    } 
} 

现在既然所有的行都包含相同数量的元素，它们可能是相关的？所以一个数据集就像用户一样，用声誉，代码和xy引用来描述一些东西。在面向对象的领域，这看起来像一个对象 - 让我们称它为用户:(如果你知道C，它就像一个结构）。

我们写一个甜美，小类：

// if we don't make it public, we can integrate it 
// into the same file. Normally, we would make it public. 

class User { 
    String name; 
    int rep; 
    char code; 
    double quote; 

    public String toString() 
    { 
     return name + "\t" + rep + "\t" + code + "\t" + quote; 
    } 

    // a constructor 
    public User (String name, int rep, char code, double quote) 
    { 
     this.name = name; 
     this.rep = rep; 
     this.code = code; 
     this.quote = quote; 
    } 
} 

，并从主方法的末尾使用它：

User [] users = new User[size]; 
    for (int i = 0; i < size; ++i) 
    { 
     users[i] = new User (names[i], numbers[i], letters[i], num2[i]); 
    } 
      // simplified for-loop and calling toString of User implicitly: 
    for (User u: users) 
     System.out.println (u); 

我不建议使用数组。一个ArrayList会更容易处理，但初学者往往被绑定到他们刚刚学到的东西，并且由于您使用Array标记了您的问题...