Question

如何打开.txt文件并将输入或空格分隔的数字读入数组列表？

Answer 1

读取文件，将每一行解析为一个整数并存储到列表中：

List<Integer> list = new ArrayList<Integer>();
File file = new File("file.txt");
BufferedReader reader = null;

try {
    reader = new BufferedReader(new FileReader(file));
    String text = null;

    while ((text = reader.readLine()) != null) {
        list.add(Integer.parseInt(text));
    }
} catch (FileNotFoundException e) {
    e.printStackTrace();
} catch (IOException e) {
    e.printStackTrace();
} finally {
    try {
        if (reader != null) {
            reader.close();
        }
    } catch (IOException e) {
    }
}

//print out the list
System.out.println(list);

Answer 2

下面是一个更短的选择：

Path filePath = Paths.get("file.txt");
Scanner scanner = new Scanner(filePath);
List<Integer> integers = new ArrayList<>();
while (scanner.hasNext()) {
    if (scanner.hasNextInt()) {
        integers.add(scanner.nextInt());
    } else {
        scanner.next();
    }
}

扫描程序使用分隔符模式将其输入分解为标记，分隔符模式默认匹配空格。虽然默认分隔符是空格，但它成功找到了由换行符分隔的所有整数。

Answer 3

Java 8中的好消息我们可以在一行中完成：

List<Integer> ints = Files.lines(Paths.get(fileName))
                          .map(Integer::parseInt)
                          .collect(Collectors.toList());

Answer 4

   try{

    BufferedReader br = new BufferedReader(new FileReader("textfile.txt"));
    String strLine;
    //Read File Line By Line
    while ((strLine = br.readLine()) != null)   {
      // Print the content on the console
      System.out.println (strLine);
    }
    //Close the input stream
    in.close();
    }catch (Exception e){//Catch exception if any
      System.err.println("Error: " + e.getMessage());
    }finally{
     in.close();
    }

这将逐行阅读，

如果你没有。由换行符char分隔。然后代替

 System.out.println (strLine);

你可以拥有

try{
int i = Integer.parseInt(strLine);
}catch(NumberFormatException npe){
//do something
}

如果用空格分隔，那么

try{
    String noInStringArr[] = strLine.split(" ");
//then you can parse it to Int as above
    }catch(NumberFormatException npe){
    //do something
    }

Answer 5

File file = new File("file.txt");   
Scanner scanner = new Scanner(file);
List<Integer> integers = new ArrayList<>();
while (scanner.hasNext()) {
    if (scanner.hasNextInt()) {
        integers.add(scanner.nextInt());
    } 
    else {
        scanner.next();
    }
}
System.out.println(integers);

Answer 6

import java.io.*;  
public class DataStreamExample {  
     public static void main(String args[]){    
          try{    
            FileWriter  fin=new FileWriter("testout.txt");    
            BufferedWriter d = new BufferedWriter(fin);
            int a[] = new int[3];
            a[0]=1;
            a[1]=22;
            a[2]=3;
            String s="";
            for(int i=0;i<3;i++)
            {
                s=Integer.toString(a[i]);
                d.write(s);
                d.newLine();
            }

            System.out.println("Success");
            d.close();
            fin.close();    



            FileReader in=new FileReader("testout.txt");
            BufferedReader br=new BufferedReader(in);
            String i="";
            int sum=0;
            while ((i=br.readLine())!= null)
            {
                sum += Integer.parseInt(i);
            }
            System.out.println(sum);
          }catch(Exception e){System.out.println(e);}    
         }    
        }

输出：：成功 26

此外，我使用数组使其变得简单...。您可以直接采用整数输入并将其转换为字符串并将其发送到文件。输入转换写入过程...就这么简单。

如何在Java中打开txt文件并读取数字

6 个答案: