Java.util.zip添加新文件会覆盖整个jar？ [英] Java.util.zip adding a new file overwrites entire jar?

问题描述

我使用java.util.zip将一些配置资源添加到jar文件中。
当我调用addFileToZip（）方法，它完全覆盖jar，而不是将文件添加到jar。为什么我需要写配置到jar是完全不相关的。并且我不想使用任何外部API。

编辑：jar不是在VM中运行，org.cfg.resource是我试图的包保存文件，文件是标准文本文档，正在编辑的jar包含在使用此方法之前的正确信息。

我的代码：

  public void addFileToZip（文件fileToAdd，文件zipFile）
 {
 ZipOutputStream zos = null; 
 FileInputStream fis = null; 
 ZipEntry ze = null; 
 byte [] buffer = null; 
 int len; 
 
 try {
 zos = new ZipOutputStream（new FileOutputStream（zipFile））; 
} catch（FileNotFoundException e）{
} 
 
 ze = new ZipEntry（org+ File.separator +cfg+ 
 File.separator + resource+ File.separator + fileToAdd.getName（））; 
 try {
 zos.putNextEntry（ze）; 
 
 fis = new FileInputStream（fileToAdd）; 
 buffer = new byte [（int）fileToAdd.length（）]; 
 
 while（（len = fis.read（buffer））> 0）
 {
 zos.write（buffer，0，len）; 
} 
} catch（IOException e）{
} 
 try {
 zos.flush（）; 
 zos.close（）; 
 fis.close（）; 
} catch（IOException e）{
} 
} 
  

解决方案

您显示的代码覆盖一个文件，无论它是否是一个zip文件。 ZipOutputStream 不关心现有数据。

我会推荐

1. 使用 ZipInputStream 创建新文件。




2. c $ c>




3. 将现有条目复制到新文件。




4. 添加新条目。

p>





---

希望在Java 7中有 Zip文件系统，这将为您节省大量工作。

我们可以直接写入zip文件中的文件

  Map< String，String> ; env = new HashMap<>（）; 
 env.put（create，true）; 
路径path = Paths.get（test.zip）; 
 URI uri = URI.create（jar：+ path.toUri（））; 
 try（FileSystem fs = FileSystems.newFileSystem（uri，env））
 {
路径nf = fs.getPath（new.txt）; 
 try（Writer writer = Files.newBufferedWriter（nf，StandardCharsets.UTF_8，StandardOpenOption.CREATE））{
 writer.write（hello）; 
} 
} 
  

I am using java.util.zip to add some configuration resources into a jar file. when I call addFileToZip() method it overwrites the jar completely, instead of adding the file to the jar. Why I need to write the config to the jar is completely irrelevant. and I do not wish to use any external API's.

EDIT: The jar is not running in the VM and org.cfg.resource is the package I'm trying to save the file to, the file is a standard text document and the jar being edited contains the proper information before this method is used.

My code:

public void addFileToZip(File fileToAdd, File zipFile)
{
    ZipOutputStream zos = null;
    FileInputStream fis = null;
    ZipEntry ze = null;
    byte[] buffer = null;
    int len;

    try {
        zos = new ZipOutputStream(new FileOutputStream(zipFile));
    } catch (FileNotFoundException e) {
    }

    ze = new ZipEntry("org" + File.separator + "cfg" + 
            File.separator + "resource" + File.separator + fileToAdd.getName());
    try {
        zos.putNextEntry(ze);

        fis = new FileInputStream(fileToAdd);
        buffer = new byte[(int) fileToAdd.length()];

        while((len = fis.read(buffer)) > 0)
        {
            zos.write(buffer, 0, len);
        }           
    } catch (IOException e) {
    }
    try {
        zos.flush();
        zos.close();
        fis.close();
    } catch (IOException e) {
    }
}

解决方案

The code you showed overrides a file no matter if it would be a zip file or not. ZipOutputStream does not care about existing data. Neither any stream oriented API does.

I would recommend

1. Create new file using ZipOutputStream.

2. Open existing with ZipInputStream

3. Copy existing entries to new file.

4. Add new entries.

5. Replace old file with a new one.

---

Hopefully in Java 7 we got Zip File System that will save you a lot of work.

We can directly write to files inside zip files

Map<String, String> env = new HashMap<>(); 
env.put("create", "true");
Path path = Paths.get("test.zip");
URI uri = URI.create("jar:" + path.toUri());
try (FileSystem fs = FileSystems.newFileSystem(uri, env))
{
    Path nf = fs.getPath("new.txt");
    try (Writer writer = Files.newBufferedWriter(nf, StandardCharsets.UTF_8, StandardOpenOption.CREATE)) {
        writer.write("hello");
    }
}

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