Question

我想编写一个程序，它将获取文件夹中存在的所有Zip文件并解压缩到目标文件夹中。我能够编写一个程序，我可以解压缩一个zip文件，但我想解压缩该文件夹中的所有zip文件，我该怎么办？

Answer 1

它不漂亮，但你明白了。

使用来自Java 7的NIO Files api流式传输zip文件的目录
使用ZIP api访问存档中的每个ZipEntry

使用NIO api将文件写入指定目录

public class Unzipper {

public static void main(String [] args){
    Unzipper unzipper = new Unzipper();
    unzipper.unzipZipsInDirTo(Paths.get("D:/"), Paths.get("D:/unzipped"));
}

public void unzipZipsInDirTo(Path searchDir, Path unzipTo ){

    final PathMatcher matcher = searchDir.getFileSystem().getPathMatcher("glob:**/*.zip");
    try (final Stream<Path> stream = Files.list(searchDir)) {
        stream.filter(matcher::matches)
                .forEach(zipFile -> unzip(zipFile,unzipTo));
    }catch (IOException e){
        //handle your exception
    }
}

public void unzip(Path zipFile, Path outputPath){
    try (ZipInputStream zis = new ZipInputStream(Files.newInputStream(zipFile))) {

        ZipEntry entry = zis.getNextEntry();

        while (entry != null) {

            Path newFilePath = outputPath.resolve(entry.getName());
            if (entry.isDirectory()) {
                Files.createDirectories(newFilePath);
            } else {
                if(!Files.exists(newFilePath.getParent())) {
                    Files.createDirectories(newFilePath.getParent());
                }
                try (OutputStream bos = Files.newOutputStream(outputPath.resolve(newFilePath))) {
                    byte[] buffer = new byte[Math.toIntExact(entry.getSize())];

                    int location;

                    while ((location = zis.read(buffer)) != -1) {
                        bos.write(buffer, 0, location);
                    }
                }
            }
            entry = zis.getNextEntry();
        }
    }catch(IOException e){
        throw new RuntimeException(e);
        //handle your exception
    }
}

}

Answer 2

您可以使用Java 7 Files API

Files.list(Paths.get("/path/to/folder"))
    .filter(c -> c.endsWith(".zip"))
    .forEach(c -> unzip(c));

如何在Java中解压缩目录中的所有Zip文件夹？

2 个答案: