从jar中读取资源文件

#1 楼

 InputStream in = getClass().getResourceAsStream("/file.txt"); 
BufferedReader reader = new BufferedReader(new InputStreamReader(in));
  
 
只要file.txt资源在类路径上可用，那么无论file.txt资源位于classes/目录中还是位于jar内部，此方法都将以相同的方式工作。
 
之所以发生URI is not hierarchical，是因为jar文件中资源的URI看起来像这样：file:/example.jar!/file.txt。您无法读取jar（zip文件）中的条目，就像它是一个普通的旧文件一样。
 
以下答案可以很好地解释这一点：
 
 
如何从Java jar文件读取资源文件？
 Java Jar文件：使用资源错误：URI不是分层的
 

评论
谢谢您，这非常有帮助并且代码可以完美地工作，但是我确实有一个问题，我需要确定InputStream是否存在（例如File.exists（）），以便我的游戏可以判断是否使用默认文件。谢谢。



 – PrinceCJC

 2013年12月5日15:23 
哦，顺便说一句，getClass（）。getResource（“ ** / folder ** / file.txt”）起作用的原因是因为我将该文件夹放在jar的同一目录中：)。



 – PrinceCJC

 2013年12月5日15:33



 
如果资源不存在，则getResourceAsStream返回null，这可以成为您的“存在”测试。



 –提请MacInnis

 2013年12月5日19:05



 
当然...不要忘记关闭inputStream和BufferedReader



 – Noremac

 15年5月15日在13:38 
我这样做了，然后将声音文件放在NetBeansProjects \ Test \ build \ classes路径中，但是使用clean和build将从该路径中清除。并且通过转换为exe，它不再有权访问声音文件。我的问题在哪里？



 –巴黎

 18 Mar 12 '18 at 19:05 

#2 楼

#3 楼

public class MyClass {

    public static InputStream accessFile() {
        String resource = "my-file-located-in-resources.txt";

        // this is the path within the jar file
        InputStream input = MyClass.class.getResourceAsStream("/resources/" + resource);
        if (input == null) {
            // this is how we load file within editor (eg eclipse)
            input = MyClass.class.getClassLoader().getResourceAsStream(resource);
        }

        return input;
    }
}

#4 楼

try {
    // Get all the files under this inner resource folder: my_folder
    String scannedPackage = "my_folder/*";
    PathMatchingResourcePatternResolver scanner = new PathMatchingResourcePatternResolver();
    Resource[] resources = scanner.getResources(scannedPackage);

    if (resources == null || resources.length == 0)
        log.warn("Warning: could not find any resources in this scanned package: " + scannedPackage);
    else {
        for (Resource resource : resources) {
            log.info(resource.getFilename());
            // Read the file content (I used BufferedReader, but there are other solutions for that):
            BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(resource.getInputStream()));
            String line = null;
            while ((line = bufferedReader.readLine()) != null) {
                // ...
                // ...                      
            }
            bufferedReader.close();
        }
    }
} catch (Exception e) {
    throw new Exception("Failed to read the resources folder: " + e.getMessage(), e);
}

#5 楼

    private List<String> decomposePath(String path){
        List<String> reversed = Lists.newArrayList();
        File currFile = new File(path);
        while(currFile != null){
            reversed.add(currFile.getName());
            currFile = currFile.getParentFile();
        }
        return Lists.reverse(reversed);
    }

    private String writeResourceToFile(String resourceName) throws IOException {
        ClassLoader loader = getClass().getClassLoader();
        InputStream configStream = loader.getResourceAsStream(resourceName);
        List<String> pathComponents = decomposePath(resourceName);
        folder.newFolder(pathComponents.subList(0, pathComponents.size() - 1).toArray(new String[0]));
        File tmpFile = folder.newFile(resourceName);
        Files.copy(configStream, tmpFile.toPath(), REPLACE_EXISTING);
        return tmpFile.getAbsolutePath();
    }

#6 楼

import java.lang.Thread;
import java.io.BufferedReader;
import java.io.InputStreamReader;

final BufferedReader in = new BufferedReader(new InputStreamReader(
      Thread.currentThread().getContextClassLoader().getResourceAsStream("file.txt"))
); // no initial slash in file.txt

#7 楼

BufferedReader br = new BufferedReader(new InputStreamReader(Main.class.getResourceAsStream(path)));

#8 楼

#9 楼

val resource = Thread.currentThread().contextClassLoader.getResource('resources.txt')

#10 楼

private String writeResourceToFile(String resourceName) throws IOException {
    File outFile = new File(certPath + File.separator + resourceName);

    if (outFile.isFile())
        return outFile.getAbsolutePath();
    
    InputStream resourceStream = null;
    
    // Java: In caso di JAR dentro il JAR applicativo 
    URLClassLoader urlClassLoader = (URLClassLoader)Cypher.class.getClassLoader();
    URL url = urlClassLoader.findResource(resourceName);
    if (url != null) {
        URLConnection conn = url.openConnection();
        if (conn != null) {
            resourceStream = conn.getInputStream();
        }
    }
    
    if (resourceStream != null) {
        Files.copy(resourceStream, outFile.toPath(), StandardCopyOption.REPLACE_EXISTING);
        return outFile.getAbsolutePath();
    } else {
        System.out.println("Embedded Resource " + resourceName + " not found.");
    }
    
    return "";
}   

#11 楼

InputStream in = getClass().getClassLoader().getResourceAsStream("file.txt"); 
BufferedReader reader = new BufferedReader(new InputStreamReader(in));

#12 楼

#13 楼

    ClassLoader classLoader = getClass().getClassLoader();
    File file = new File(classLoader.getResource("file/test.xml").getFile());

#14 楼

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import org.springframework.core.io.ClassPathResource;

  public String readFileToString(String path) throws IOException {

    StringBuilder resultBuilder = new StringBuilder("");
    ClassPathResource resource = new ClassPathResource(path);

    try (
        InputStream inputStream = resource.getInputStream();
        BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(inputStream))) {

      String line;

      while ((line = bufferedReader.readLine()) != null) {
        resultBuilder.append(line);
      }

    }

    return resultBuilder.toString();
  }

#15 楼

val authReader = InputStreamReader(javaClass.getResourceAsStream("/file1.json"))