如何将jar中的文件复制到jar外?

时间:2012-04-25 01:29:24

标签: java file jar copy working-directory

我想从jar中复制一个文件。我要复制的文件将被复制到工作目录之外。我做了一些测试,我尝试的所有方法都以0字节文件结束。

编辑:我希望通过程序复制文件,而不是手动复制。

9 个答案:

答案 0 :(得分:48)

首先,我想说之前发布的一些答案是完全正确的,但我想给我的,因为有时我们不能在GPL下使用开源库,或者因为我们懒得下载jar XD或者你的理由是一个独立的解决方案。

下面的函数复制Jar文件旁边的资源:

  /**
     * Export a resource embedded into a Jar file to the local file path.
     *
     * @param resourceName ie.: "/SmartLibrary.dll"
     * @return The path to the exported resource
     * @throws Exception
     */
    static public String ExportResource(String resourceName) throws Exception {
        InputStream stream = null;
        OutputStream resStreamOut = null;
        String jarFolder;
        try {
            stream = ExecutingClass.class.getResourceAsStream(resourceName);//note that each / is a directory down in the "jar tree" been the jar the root of the tree
            if(stream == null) {
                throw new Exception("Cannot get resource \"" + resourceName + "\" from Jar file.");
            }

            int readBytes;
            byte[] buffer = new byte[4096];
            jarFolder = new File(ExecutingClass.class.getProtectionDomain().getCodeSource().getLocation().toURI().getPath()).getParentFile().getPath().replace('\\', '/');
            resStreamOut = new FileOutputStream(jarFolder + resourceName);
            while ((readBytes = stream.read(buffer)) > 0) {
                resStreamOut.write(buffer, 0, readBytes);
            }
        } catch (Exception ex) {
            throw ex;
        } finally {
            stream.close();
            resStreamOut.close();
        }

        return jarFolder + resourceName;
    }

只需将ExecutingClass更改为您的类的名称,并将其命名为:

String fullPath = ExportResource("/myresource.ext");

编辑Java 7+(为方便起见)

GOXR3PLUS回答并由Andy Thomas注明,您可以通过以下方式实现此目的:

Files.copy( InputStream in, Path target, CopyOption... options)

有关详细信息,请参阅GOXR3PLUS answer

答案 1 :(得分:34)

鉴于您对0字节文件的评论,我不得不假设您正在尝试以编程方式执行此操作,并且,鉴于您的标记,您正在使用Java进行此操作。如果这是真的,那么只需使用Class.getResource()获取指向JAR中文件的URL,然后使用Apache Commons IO FileUtils.copyURLToFile()将其复制到文件系统。 E.g:

URL inputUrl = getClass().getResource("/absolute/path/of/source/in/jar/file");
File dest = new File("/path/to/destination/file");
FileUtils.copyURLToFile(inputUrl, dest);

最有可能的是,您现在拥有的任何代码的问题是您(正确地)使用缓冲输出流写入文件但(错误地)无法关闭它。

哦,你应该编辑你的问题,以明确地说明 你想要做什么(以编程方式,而不是语言......)

答案 2 :(得分:11)

Java 8(实际上是自1.7以来的FileSystem)附带了一些很酷的新类/方法来处理这个问题。有人已经提到JAR基本上是ZIP文件,你可以使用

final URI jarFileUril = URI.create("jar:file:" + file.toURI().getPath());
final FileSystem fs = FileSystems.newFileSystem(jarFileUri, env);

(见Zip File

然后你可以使用一种方便的方法,如:

fs.getPath("filename");

然后你可以使用Files class

try (final Stream<Path> sources = Files.walk(from)) {
     sources.forEach(src -> {
         final Path dest = to.resolve(from.relativize(src).toString());
         try {
            if (Files.isDirectory(from)) {
               if (Files.notExists(to)) {
                   log.trace("Creating directory {}", to);
                   Files.createDirectories(to);
               }
            } else {
                log.trace("Extracting file {} to {}", from, to);
                Files.copy(from, to, StandardCopyOption.REPLACE_EXISTING);
            }
       } catch (IOException e) {
           throw new RuntimeException("Failed to unzip file.", e);
       }
     });
}

注意:我尝试解压缩JAR文件以进行测试

答案 3 :(得分:10)

使用 Java 7 + 更快的方式,以及获取当前目录的代码:

   /**
     * Copy a file from source to destination.
     *
     * @param source
     *        the source
     * @param destination
     *        the destination
     * @return True if succeeded , False if not
     */
    public static boolean copy(InputStream source , String destination) {
        boolean succeess = true;

        System.out.println("Copying ->" + source + "\n\tto ->" + destination);

        try {
            Files.copy(source, Paths.get(destination), StandardCopyOption.REPLACE_EXISTING);
        } catch (IOException ex) {
            logger.log(Level.WARNING, "", ex);
            succeess = false;
        }

        return succeess;

    }

测试它( icon.png 是应用程序包图像中的图像):

copy(getClass().getResourceAsStream("/image/icon.png"),getBasePathForClass(Main.class)+"icon.png");

关于代码行(getBasePathForClass(Main.class)): - &gt;检查我在这里添加的答案:) - &gt; Getting the Current Working Directory in Java

答案 4 :(得分:1)

使用JarInputStream类:

// assuming you already have an InputStream to the jar file..
JarInputStream jis = new JarInputStream( is );
// get the first entry
JarEntry entry = jis.getNextEntry();
// we will loop through all the entries in the jar file
while ( entry != null ) {
  // test the entry.getName() against whatever you are looking for, etc
  if ( matches ) {
    // read from the JarInputStream until the read method returns -1
    // ...
    // do what ever you want with the read output
    // ...
    // if you only care about one file, break here 
  }
  // get the next entry
  entry = jis.getNextEntry();
}
jis.close();

另请参阅:JarEntry

答案 5 :(得分:1)

强大的解决方案:

public static void copyResource(String res, String dest, Class c) throws IOException {
    InputStream src = c.getResourceAsStream(res);
    Files.copy(src, Paths.get(dest), StandardCopyOption.REPLACE_EXISTING);
}

您可以像这样使用它:

File tempFileGdalZip = File.createTempFile("temp_gdal", ".zip");
copyResource("/gdal.zip", tempFileGdalZip.getAbsolutePath(), this.getClass());

答案 6 :(得分:0)

要将文件从jar复制到外部,您需要使用以下方法:

  1. 使用getResourceAsStream()
  2. 为您的jar文件中的文件获取InputStream
  3. 我们使用FileOutputStream
  4. 打开目标文件
  5. 我们将字节从输入复制到输出流
  6. 我们关闭流以防止资源泄漏
  7. 示例代码,其中还包含一个不替换现有值的变量:

    public File saveResource(String name) throws IOException {
        return saveResource(name, true);
    }
    
    public File saveResource(String name, boolean replace) throws IOException {
        return saveResource(new File("."), name, replace)
    }
    
    public File saveResource(File outputDirectory, String name) throws IOException {
        return saveResource(outputDirectory, name, true);
    }
    
    public File saveResource(File outputDirectory, String name, boolean replace)
           throws IOException {
        File out = new File(outputDirectory, name);
        if (!replace && out.exists()) 
            return out;
        // Step 1:
        InputStream resource = this.getClass().getResourceAsStream(name);
        if (resource == null)
           throw new FileNotFoundException(name + " (resource not found)");
        // Step 2 and automatic step 4
        try(InputStream in = resource;
            OutputStream writer = new BufferedOutputStream(
                new FileOutputStream(out))) {
             // Step 3
             byte[] buffer = new byte[1024 * 4];
             int length;
             while((length = in.read(buffer)) >= 0) {
                 writer.write(buffer, 0, length);
             }
         }
         return out;
    }
    

答案 7 :(得分:-1)

jar只是一个zip文件。解压缩(使用您熟悉的任何方法)并正常复制文件。

答案 8 :(得分:-2)

${JAVA_HOME}/bin/jar -cvf /path/to.jar