在Java中从文件中读取数据的最简单方法是使用file类读取文件，使用Scanner类读取文件的内容。

public static void main(String args[])throws Exception
{
   File f = new File("input.txt");
   takeInputIn2DArray(f);
}

public static void takeInputIn2DArray(File f) throws Exception
{
    Scanner s = new Scanner(f);
    int a[][] = new int[20][20];
    for(int i=0; i<20; i++)
    {
        for(int j=0; j<20; j++)
        {
            a[i][j] = s.nextInt();
        }
    }
}

PS:别忘了导入java.util.*;扫描仪的工作。

可能没有缓冲I/O那么快，但是非常简洁:

    String content;
    try (Scanner scanner = new Scanner(textFile).useDelimiter("\\Z")) {
        content = scanner.next();
    }

\Z模式告诉扫描器分隔符是EOF。

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

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); }); } } }

如果您有一个大文件，您可以使用Apache Commons IO迭代处理该文件，而不会耗尽可用内存。

try (LineIterator it = FileUtils.lineIterator(theFile, "UTF-8")) {
    while (it.hasNext()) {
        String line = it.nextLine();
        // do something with line
    }
}

对于基于jsf的Maven web应用程序，只需使用ClassLoader和Resources文件夹读取任何你想要的文件:

Put any file you want to read in the Resources folder. Put the Apache Commons IO dependency into your POM: <dependency> <groupId>org.apache.commons</groupId> <artifactId>commons-io</artifactId> <version>1.3.2</version> </dependency> Use the code below to read it (e.g. below is reading in a .json file): String metadata = null; FileInputStream inputStream; try { ClassLoader loader = Thread.currentThread().getContextClassLoader(); inputStream = (FileInputStream) loader .getResourceAsStream("/metadata.json"); metadata = IOUtils.toString(inputStream); inputStream.close(); } catch (FileNotFoundException e) { // TODO Auto-generated catch block e.printStackTrace(); } catch (IOException e) { // TODO Auto-generated catch block e.printStackTrace(); } return metadata;

您可以对文本文件、.properties文件、XSD模式等执行相同的操作。