如何读取没有清单的war/jar文件

发布于 2024-11-06 07:57:19 字数 701 浏览 0 评论 0原文

我有一个 war 文件，它不包含清单，甚至不包含 META-INF 文件夹。现在我的问题是我编写了一个代码，该代码可以很好地处理包含清单的普通战争文件。现在我需要读取一个不包含清单的战争文件。

当我检查

while ((ze = zis.getNextEntry()) != null)

此条件时，会被跳过。是否有任何 API 将其视为普通的 zip 文件，或者是否有任何解决方法。

我尝试过使用 JarEntry 以及 ZipEntry。这是一个应该可以解释的小片段。

try {
            FileInputStream fis = new FileInputStream(applicationPack);
            ZipArchiveInputStream zis = new ZipArchiveInputStream(fis);
            ArchiveEntry ze = null;
            File applicationPackConfiguration;           
            while ((ze = zis.getNextEntry()) != null) {
            // do someting
}

可以做什么？

原文

I have a war file which does not contains manifest not even META-INF folder. Now my problem is that I wrote a code which was working fine with normal war files containing manifests. Now I am required to read a war file which does not contain manifest.

When I check

while ((ze = zis.getNextEntry()) != null)

This condition is just skipped. Is there any API which treats it just as a normal zip file or is there any workaround.

I have tried with JarEntry as well as ZipEntry. Here is a small snippet that should be explanatory.

try {
            FileInputStream fis = new FileInputStream(applicationPack);
            ZipArchiveInputStream zis = new ZipArchiveInputStream(fis);
            ArchiveEntry ze = null;
            File applicationPackConfiguration;           
            while ((ze = zis.getNextEntry()) != null) {
            // do someting
}

What can be done ?

分享到QQ

分享到微博

如果你对这篇内容有疑问，欢迎到本站社区发帖提问参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。

发布评论

需要登录才能够评论，你可以免费注册一个本站的账号。

-黛色若梦 2024-11-13 07:57:19

您可以简单地使用 ZipFile 类列出内容：

try {
  // Open the ZIP file
  ZipFile zf = new ZipFile("filename.zip");

  // Enumerate each entry
  for (Enumeration entries = zf.entries(); entries.hasMoreElements();) {
    // Get the entry name
    String zipEntryName = ((ZipEntry)entries.nextElement()).getName();
  }
} catch (IOException e) {
}

示例取自 < a href="http://www.exampledepot.com/egs/java.util.zip/ListZip.html" rel="nofollow">此处。从 zip 检索文件的另一个示例。

更新：

上面的代码确实对仅包含目录作为顶级元素的 zip 文件存在问题。

此代码有效（已测试）：

    try {
        // Open the ZIP file
        FileInputStream fis = new FileInputStream(new File("/your.war"));
        ZipInputStream zis = new ZipInputStream(new BufferedInputStream(fis));

        ZipEntry entry = null;

        while ((entry = zis.getNextEntry()) != null)
            // Get the entry name
            System.out.println(entry.getName());
    } catch (IOException e) {
    }

You can simply list contents with ZipFile class:

try {
  // Open the ZIP file
  ZipFile zf = new ZipFile("filename.zip");

  // Enumerate each entry
  for (Enumeration entries = zf.entries(); entries.hasMoreElements();) {
    // Get the entry name
    String zipEntryName = ((ZipEntry)entries.nextElement()).getName();
  }
} catch (IOException e) {
}

Example taken from here. Another example for retrieving the file from zip.

Update:

Code above indeed has problems with zip files that contain only directory as a top-level element.

This code works (tested):

    try {
        // Open the ZIP file
        FileInputStream fis = new FileInputStream(new File("/your.war"));
        ZipInputStream zis = new ZipInputStream(new BufferedInputStream(fis));

        ZipEntry entry = null;

        while ((entry = zis.getNextEntry()) != null)
            // Get the entry name
            System.out.println(entry.getName());
    } catch (IOException e) {
    }

回复收藏 0 原文

放飞的风筝 2024-11-13 07:57:19

您可以使用 java.lang.Class 中的类。 util.zip 包。只需将 ZipArchiveInputStream 替换为 ZipInputStream，将 ArchiveEntry 替换为 ZipEntry：

FileInputStream fis = new FileInputStream(new File("/path/to/your.war"));
ZipInputStream zis = new ZipInputStream(fis);
ZipEntry ze = null;
while ((ze = zis.getNextEntry()) != null) {
   System.out.println(ze.getName());
}

You can use classes from java.util.zip package. Just replace ZipArchiveInputStream with ZipInputStream and ArchiveEntry with ZipEntry:

FileInputStream fis = new FileInputStream(new File("/path/to/your.war"));
ZipInputStream zis = new ZipInputStream(fis);
ZipEntry ze = null;
while ((ze = zis.getNextEntry()) != null) {
   System.out.println(ze.getName());
}

回复收藏 0 原文

~没有更多了~