您说“我正在尝试读取一个设置在CLASSPATH系统变量中的文本文件。”我猜这是在Windows上，你正在使用这个丑陋的对话框编辑“系统变量”。

现在在控制台中运行Java程序。但这并不管用:控制台在启动时只获得系统变量值的副本。这意味着之后对话中的任何改变都不会产生任何影响。

有这些解决方案:

每次更改后启动一个新控制台 使用set CLASSPATH=…在控制台中设置变量的副本，当您的代码工作时，将最后一个值粘贴到变量对话框中。 将对Java的调用放入. bat文件并双击它。这将每次创建一个新的控制台(因此复制系统变量的当前值)。

注意:如果你也有一个用户变量CLASSPATH，那么它将遮蔽你的系统变量。这就是为什么将对Java程序的调用放到. bat文件中并在其中设置类路径(使用set classpath =)而不是依赖于全局系统或用户变量的原因。

这也确保您的计算机上可以有多个Java程序，因为它们必然具有不同的类路径。

通过类路径上的目录，从同一个类加载器加载的类中，你应该能够使用以下任何一种:

// From ClassLoader, all paths are "absolute" already - there's no context
// from which they could be relative. Therefore you don't need a leading slash.
InputStream in = this.getClass().getClassLoader()
                                .getResourceAsStream("SomeTextFile.txt");
// From Class, the path is relative to the package of the class unless
// you include a leading slash, so if you don't want to use the current
// package, include a slash like this:
InputStream in = this.getClass().getResourceAsStream("/SomeTextFile.txt");

如果这些都不起作用，那就说明有其他问题。

举个例子，这段代码:

package dummy;

import java.io.*;

public class Test
{
    public static void main(String[] args)
    {
        InputStream stream = Test.class.getResourceAsStream("/SomeTextFile.txt");
        System.out.println(stream != null);
        stream = Test.class.getClassLoader().getResourceAsStream("SomeTextFile.txt");
        System.out.println(stream != null);
    }
}

这个目录结构是:

code
    dummy
          Test.class
txt
    SomeTextFile.txt

然后(使用Unix路径分隔符，因为我在Linux机器上):

java -classpath code:txt dummy.Test

结果:

true
true

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

public class ReadFile

{
    /**
     * * feel free to make any modification I have have been here so I feel you
     * * * @param args * @throws InterruptedException
     */

    public static void main(String[] args) throws InterruptedException {
        // thread pool of 10
        File dir = new File(".");
        // read file from same directory as source //
        if (dir.isDirectory()) {
            File[] files = dir.listFiles();
            for (File file : files) {
                // if you wanna read file name with txt files
                if (file.getName().contains("txt")) {
                    System.out.println(file.getName());
                }

                // if you want to open text file and read each line then
                if (file.getName().contains("txt")) {
                    try {
                        // FileReader reads text files in the default encoding.
                        FileReader fileReader = new FileReader(
                                file.getAbsolutePath());
                        // Always wrap FileReader in BufferedReader.
                        BufferedReader bufferedReader = new BufferedReader(
                                fileReader);
                        String line;
                        // get file details and get info you need.
                        while ((line = bufferedReader.readLine()) != null) {
                            System.out.println(line);
                            // here you can say...
                            // System.out.println(line.substring(0, 10)); this
                            // prints from 0 to 10 indext
                        }
                    } catch (FileNotFoundException ex) {
                        System.out.println("Unable to open file '"
                                + file.getName() + "'");
                    } catch (IOException ex) {
                        System.out.println("Error reading file '"
                                + file.getName() + "'");
                        // Or we could just do this:
                        ex.printStackTrace();
                    }
                }
            }
        }

    }

}

我的回答与问题中所问的不完全一样。相反，我给出了一个解决方案，我们可以很容易地从我们的项目类路径读取文件到java应用程序。

例如，假设配置文件example.xml位于如下路径

com.myproject.config.dev

我们的Java可执行类文件在下面的路径:-

com.myproject.server.main

now just check in both the above path which is the nearest common directory/folder from where you can access both dev and main directory/folder (com.myproject.server.main - where our application’s java executable class is existed) – We can see that it is myproject folder/directory which is the nearest common directory/folder from where we can access our example.xml file. Therefore from a java executable class resides in folder/directory main we have to go back two steps like ../../ to access myproject. Now following this, see how we can read the file:-

package com.myproject.server.main;

class Example {

  File xmlFile;

  public Example(){
       String filePath = this.getClass().getResource("../../config/dev/example.xml").getPath();
       this.xmlFile = new File(filePath);
    }

  public File getXMLFile() {
      return this.xmlFile;
  }
   public static void main(String args[]){
      Example ex = new Example();
      File xmlFile = ex.getXMLFile();
   }
}

不要使用getClassLoader()方法，并在文件名前使用“/”。“/”非常重要

this.getClass().getResourceAsStream("/SomeTextFile.txt");

使用org.apache.commons.io.FileUtils.readFileToString(新文件("src/test/resources/sample-data/fileName.txt"));