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); }); } } }

如果这是关于结构的简单性，请使用Java kiss:

import static kiss.API.*;

class App {
  void run() {
    String line;
    try (Close in = inOpen("file.dat")) {
      while ((line = readLine()) != null) {
        println(line);
      }
    }
  }
}

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

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。

最简单的方法是使用Java中的Scanner类和FileReader对象。简单的例子:

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

扫描器有几个方法读取字符串，数字，等…您可以在Java文档页面上查找有关这方面的更多信息。

例如，将整个内容读入String:

StringBuilder sb = new StringBuilder();
while(in.hasNext()) {
    sb.append(in.next());
}
in.close();
outString = sb.toString();

另外，如果你需要一个特定的编码，你可以使用这个来代替FileReader:

new InputStreamReader(new FileInputStream(fileUtf8), StandardCharsets.UTF_8)

org.apache.commons.io.FileUtils中的方法也可能非常方便，例如:

/**
 * Reads the contents of a file line by line to a List
 * of Strings using the default encoding for the VM.
 */
static List readLines(File file)