如何在 memory 中提取 zip 文件

Question

我已经阅读了许多将 zip 文件提取到磁盘上的主题。 但是我有一个用例，需要在 memory 中提取 zip。 ZIP 文件再次包含 zip 文件的列表。 在阅读了堆栈溢出中的几篇文章后，我问了这个问题。 您能否分享任何有关如何在 memory 中解压缩文件的信息的帖子/链接？

Answer 1

如果您想读取nested.zip 文件，您可以尝试使用ZipInputStream （就像已经提到的那样）并检查ZipEntry (s) 是否也是 *.zip 文件，在这种情况下可以递归读取为 next.ZADCDBD79A8D84025 文件. 就像是：

private static void readZipInputStream(
        InputStream inputStream, BiConsumer<ZipEntry, ByteArrayOutputStream> consumerFunction) throws IOException {

    try (ZipInputStream zipInput = new ZipInputStream(inputStream)) {
        ZipEntry entry;
        while ((entry = zipInput.getNextEntry()) != null) {
            ByteArrayOutputStream outStream = new ByteArrayOutputStream();
            byte[] buffer = new byte[1024];
            int length;
            while ((length = zipInput.read(buffer)) != -1) {
                outStream.write(buffer, 0, length);
            }

            if (entry.getName().endsWith(".zip")) {
                // need to go deeper...
                ByteArrayInputStream inStream = new ByteArrayInputStream(outStream.toByteArray());
                readZipInputStream(inStream, consumerFunction);
            } else {
                // do something...
                consumerFunction.accept(entry, outStream);
            }
        }
    }
}

例如，有一个 zip 文件，其结构如下：

file.zip
├─1+2.zip
│ ├─1.zip
│ │ └─1.txt
│ └─2.zip
│   └─2.txt
└─3.zip
  └─3.txt

并像这样使用readZipInputStream function ：

public class Application {

    public static void main(String[] args) throws IOException {
        String path = "file.zip";
        try (FileInputStream inputStream = new FileInputStream(Paths.get(path).toFile())) {
            readZipInputStream(
                    inputStream,
                    (entry, outputStream) -> {
                        System.out.println(entry.getName());
                        System.out.println("--------------------------------");
                        System.out.println(outputStream.toString());
                        System.out.println("--------------------------------");
                    }
            );
        }
    }
}

将打印三个.txt 文件的内容：

1.txt
--------------------------------
Lorem ipsum dolor sit amet, consectetur adipisicing elit, sed do eiusmod
tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam,
quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo
consequat.
--------------------------------
2.txt
--------------------------------
- Integer vel sem consectetur, ullamcorper leo quis, consequat mauris.
- Nulla efficitur sapien at velit fermentum condimentum.
- Vestibulum elementum nulla ut ipsum tempus, ut molestie sem sollicitudin.
--------------------------------
3.txt
--------------------------------
Morbi tincidunt ornare mi. Sed id risus tortor. Interdum et malesuada 
fames ac ante ipsum primis in faucibus. Pellentesque tincidunt, 
nulla a interdum porta, orci elit ultricies leo, in maximus orci 
tortor pulvinar est. Curabitur eget fermentum risus. Vestibulum euismod 
convallis eros, nec blandit neque blandit at.
--------------------------------

Answer 2

The java class java.util.zip.ZipInputStream allows you to read the data from a Zip archive into a byte array.