Question

我正在制作一款需要更新的游戏。

我有两个JAR文件：Update.Jar和Game.Jar 基本上，我希望能够修改Game.Jar而不会完全覆盖它。

我想：

从代码
替换/添加一些资源
重新包装Jar文件。

有没有简单的方法或类可以做到这一点？如果没有，那么最干净的方法是什么？

Answer 1

Java JAR文件是普通的ZIP文件。因此，您可以使用处理ZIP的代码打开并修改它。

这是一个有效的片段（由大卫提供）：

import java.io.File;
import java.io.FileInputStream;
import java.io.FileOutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.util.ArrayList;
import java.util.zip.ZipEntry;
import java.util.zip.ZipInputStream;
import java.util.zip.ZipOutputStream;


public class JarUpdater {
    public static void main(String[] args) {

        File[] contents = {new File("F:\\ResourceTest.txt"),
                           new File("F:\\ResourceTest2.bmp")};

        File jarFile = new File("F:\\RepackMe.jar");

        try {
            updateZipFile(jarFile, contents);
        } catch (IOException e) {
            e.printStackTrace();
        }

    }

    public static void updateZipFile(File zipFile,
             File[] files) throws IOException {
               // get a temp file
        File tempFile = File.createTempFile(zipFile.getName(), null);
               // delete it, otherwise you cannot rename your existing zip to it.
        tempFile.delete();

        boolean renameOk=zipFile.renameTo(tempFile);
        if (!renameOk)
        {
            throw new RuntimeException("could not rename the file "+zipFile.getAbsolutePath()+" to "+tempFile.getAbsolutePath());
        }
        byte[] buf = new byte[1024];

        ZipInputStream zin = new ZipInputStream(new FileInputStream(tempFile));
        ZipOutputStream out = new ZipOutputStream(new FileOutputStream(zipFile));

        ZipEntry entry = zin.getNextEntry();
        while (entry != null) {
            String name = entry.getName();
            boolean notInFiles = true;
            for (File f : files) {
                if (f.getName().equals(name)) {
                    notInFiles = false;
                    break;
                }
            }
            if (notInFiles) {
                // Add ZIP entry to output stream.
                out.putNextEntry(new ZipEntry(name));
                // Transfer bytes from the ZIP file to the output file
                int len;
                while ((len = zin.read(buf)) > 0) {
                    out.write(buf, 0, len);
                }
            }
            entry = zin.getNextEntry();
        }
        // Close the streams        
        zin.close();
        // Compress the files
        for (int i = 0; i < files.length; i++) {
            InputStream in = new FileInputStream(files[i]);
            // Add ZIP entry to output stream.
            out.putNextEntry(new ZipEntry(files[i].getName()));
            // Transfer bytes from the file to the ZIP file
            int len;
            while ((len = in.read(buf)) > 0) {
                out.write(buf, 0, len);
            }
            // Complete the entry
            out.closeEntry();
            in.close();
        }
        // Complete the ZIP file
        out.close();
        tempFile.delete();
    }
}

Answer 2

在Java 7中，最好的方法是使用内置的Zip文件系统提供程序：

http://docs.oracle.com/javase/7/docs/technotes/guides/io/fsp/zipfilesystemprovider.html

即

import java.util.*;
import java.net.URI;
import java.nio.file.Path;
import java.nio.file.*;

public class ZipFSPUser {
    public static void main(String [] args) throws Throwable {
        Map<String, String> env = new HashMap<>();
        env.put("create", "true");
        // locate file system by using the syntax
        // defined in java.net.JarURLConnection
        URI uri = URI.create("jar:file:/codeSamples/zipfs/zipfstest.zip");

       try (FileSystem zipfs = FileSystems.newFileSystem(uri, env)) {
            Path externalTxtFile = Paths.get("/codeSamples/zipfs/SomeTextFile.txt");
            Path pathInZipfile = zipfs.getPath("/SomeTextFile.txt");      
            // copy a file into the zip file
            Files.copy( externalTxtFile,pathInZipfile,
                    StandardCopyOption.REPLACE_EXISTING );
        }
    }
}

〜

从代码更新.JAR的内容

2 个答案: