不知怎的，最好的答案对我不起作用。我需要使用稍微不同的代码。

ClassLoader loader = Thread.currentThread().getContextClassLoader();
InputStream is = loader.getResourceAsStream("SomeTextFile.txt");

我希望这能帮助那些遇到同样问题的人。

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

这是我如何读取我的类路径上的文本文件的所有行，使用Java 7 NIO:

...
import java.nio.charset.Charset;
import java.nio.file.Files;
import java.nio.file.Paths;

...

Files.readAllLines(
    Paths.get(this.getClass().getResource("res.txt").toURI()), Charset.defaultCharset());

注意，这是如何做到这一点的一个例子。你必须做出必要的改进。此示例仅在文件实际出现在您的类路径中时有效，否则当getResource()返回null并对其调用. touri()时将抛出NullPointerException。

另外，从Java 7开始，指定字符集的一种方便方法是使用Java .nio.charset. standardcharsets中定义的常量 (根据它们的javadocs，它们“保证在Java平台的每个实现上都可用”)。

因此，如果您知道文件的编码为UTF-8，则显式指定字符集StandardCharsets。UTF_8

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

要获得类的绝对路径，请尝试以下方法:

String url = this.getClass().getResource("").getPath();

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();
   }
}