try (Stream<String> stream = Files.lines(Paths.get(String.valueOf(new File("yourFile.txt"))))) {
    stream.forEach(System.out::println);
} catch (IOException e) {
    e.printStackTrace();
}

新文件(< path_name >)

通过将给定的路径名字符串转换为抽象路径名来创建一个新的File实例。如果给定的字符串是空字符串，那么结果就是空的抽象路径名。 参数: pathname -路径名字符串 抛出: NullPointerException -如果路径名参数为空

文件。lines返回String流

Stream<String> Stream = Files.lines(路径。get(字符串。返回对象的值(新文件(“yourFile.txt”)))) 可以抛出nullpointerexction, FileNotFoundException，所以，保持它在尝试将照顾异常在运行时

stream.forEach(System.out::println);

这用于在控制台中迭代流和打印 如果您有不同的用例，您可以提供您的自定义函数来操作行流

这基本上与Jesus Ramos的回答完全相同，除了使用File而不是FileReader加上迭代来逐级遍历文件的内容。

Scanner in = new Scanner(new File("filename.txt"));

while (in.hasNext()) { // Iterates each line in the file
    String line = in.nextLine();
    // Do something with line
}

in.close(); // Don't forget to close resource leaks

．.． 抛出FileNotFoundException

使用BufferedReader:

import java.io.BufferedReader;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;

BufferedReader br;
try {
    br = new BufferedReader(new FileReader("/fileToRead.txt"));
    try {
        String x;
        while ( (x = br.readLine()) != null ) {
            // Printing out each line in the file
            System.out.println(x);
        }
    }
    catch (IOException e) {
        e.printStackTrace();
    }
}
catch (FileNotFoundException e) {
    System.out.println(e);
    e.printStackTrace();
}

import java.util.stream.Stream;
import java.nio.file.*;
import java.io.*;

class ReadFile {

 public static void main(String[] args) {

    String filename = "Test.txt";

    try(Stream<String> stream = Files.lines(Paths.get(filename))) {

          stream.forEach(System.out:: println);

    } catch (IOException e) {

        e.printStackTrace();
    }

 }

 }

只需使用java 8流。

我记录了15种用Java读取文件的方法，然后测试了它们在不同文件大小下的速度——从1kb到1gb，下面是最常用的三种方法:

java.nio.file.Files.readAllBytes() Tested to work in Java 7, 8, and 9. import java.io.File; import java.io.IOException; import java.nio.file.Files; public class ReadFile_Files_ReadAllBytes { public static void main(String [] pArgs) throws IOException { String fileName = "c:\\temp\\sample-10KB.txt"; File file = new File(fileName); byte [] fileBytes = Files.readAllBytes(file.toPath()); char singleChar; for(byte b : fileBytes) { singleChar = (char) b; System.out.print(singleChar); } } } java.io.BufferedReader.readLine() Tested to work in Java 7, 8, 9. import java.io.BufferedReader; import java.io.FileReader; import java.io.IOException; public class ReadFile_BufferedReader_ReadLine { public static void main(String [] args) throws IOException { String fileName = "c:\\temp\\sample-10KB.txt"; FileReader fileReader = new FileReader(fileName); try (BufferedReader bufferedReader = new BufferedReader(fileReader)) { String line; while((line = bufferedReader.readLine()) != null) { System.out.println(line); } } } } java.nio.file.Files.lines() This was tested to work in Java 8 and 9 but won't work in Java 7 because of the lambda expression requirement. import java.io.File; import java.io.IOException; import java.nio.file.Files; import java.util.stream.Stream; public class ReadFile_Files_Lines { public static void main(String[] pArgs) throws IOException { String fileName = "c:\\temp\\sample-10KB.txt"; File file = new File(fileName); try (Stream linesStream = Files.lines(file.toPath())) { linesStream.forEach(line -> { System.out.println(line); }); } } }

在实践中，缓冲流类的性能要高得多，以至于NIO.2 API包含了专门返回这些流类的方法，部分原因是为了鼓励您始终在应用程序中使用缓冲流。

这里有一个例子:

Path path = Paths.get("/myfolder/myfile.ext");
try (BufferedReader reader = Files.newBufferedReader(path)) {
    // Read from the stream
    String currentLine = null;
    while ((currentLine = reader.readLine()) != null)
        //do your code here
} catch (IOException e) {
    // Handle file I/O exception...
}

您可以替换此代码

BufferedReader reader = Files.newBufferedReader(path);

与

BufferedReader br = new BufferedReader(new FileReader("/myfolder/myfile.ext"));

我推荐这篇文章来学习Java NIO和IO的主要用途。