Question

是否有任何示例代码，如何将ZIP中的文件夹解压缩到我想要的目录中？我已将文件夹“FOLDER”中的所有文件读入字节数组，如何从其文件结构中重新创建？

Answer 1

我不确定你的意思是什么？你的意思是在没有API帮助的情况下自己动手吗？

如果您不介意使用某些开源库，那里有一个很酷的API叫zip4J

它易于使用，我认为有很好的反馈。见这个例子：

String source = "folder/source.zip";
String destination = "folder/source/";   

try {
    ZipFile zipFile = new ZipFile(source);
    zipFile.extractAll(destination);
} catch (ZipException e) {
    e.printStackTrace();
}

如果您要解压缩的文件有密码，可以试试这个：

String source = "folder/source.zip";
String destination = "folder/source/";
String password = "password";

try {
    ZipFile zipFile = new ZipFile(source);
    if (zipFile.isEncrypted()) {
        zipFile.setPassword(password);
    }
    zipFile.extractAll(destination);
} catch (ZipException e) {
    e.printStackTrace();
}

我希望这很有用。

Answer 2

这是我正在使用的代码。根据您的需要更改BUFFER_SIZE。

import java.io.BufferedOutputStream;
import java.io.File;
import java.io.FileInputStream;
import java.io.FileOutputStream;
import java.io.IOException;
import java.util.zip.ZipEntry;
import java.util.zip.ZipInputStream;

public class ZipUtils
{
  private static final int  BUFFER_SIZE = 4096;

  private static void extractFile(ZipInputStream in, File outdir, String name) throws IOException
  {
    byte[] buffer = new byte[BUFFER_SIZE];
    BufferedOutputStream out = new BufferedOutputStream(new FileOutputStream(new File(outdir,name)));
    int count = -1;
    while ((count = in.read(buffer)) != -1)
      out.write(buffer, 0, count);
    out.close();
  }

  private static void mkdirs(File outdir,String path)
  {
    File d = new File(outdir, path);
    if( !d.exists() )
      d.mkdirs();
  }

  private static String dirpart(String name)
  {
    int s = name.lastIndexOf( File.separatorChar );
    return s == -1 ? null : name.substring( 0, s );
  }

  /***
   * Extract zipfile to outdir with complete directory structure
   * @param zipfile Input .zip file
   * @param outdir Output directory
   */
  public static void extract(File zipfile, File outdir)
  {
    try
    {
      ZipInputStream zin = new ZipInputStream(new FileInputStream(zipfile));
      ZipEntry entry;
      String name, dir;
      while ((entry = zin.getNextEntry()) != null)
      {
        name = entry.getName();
        if( entry.isDirectory() )
        {
          mkdirs(outdir,name);
          continue;
        }
        /* this part is necessary because file entry can come before
         * directory entry where is file located
         * i.e.:
         *   /foo/foo.txt
         *   /foo/
         */
        dir = dirpart(name);
        if( dir != null )
          mkdirs(outdir,dir);

        extractFile(zin, outdir, name);
      }
      zin.close();
    } 
    catch (IOException e)
    {
      e.printStackTrace();
    }
  }
}

Answer 3

使用Ant Compress库可以实现同样的效果。它将保留文件夹结构。

Maven依赖： -

<dependency>
    <groupId>org.apache.ant</groupId>
    <artifactId>ant-compress</artifactId>
    <version>1.2</version>
</dependency>

示例代码： -

Unzip unzipper = new Unzip();
unzipper.setSrc(theZIPFile);
unzipper.setDest(theTargetFolder);
unzipper.execute();

Answer 4

这是一个简单的解决方案，遵循更现代的惯例。如果要解压缩较大的文件，可能需要将缓冲区大小更改为更小。这样您就不会将所有文件信息保存在内存中。

    public static void unzip(File source, String out) throws IOException {
    try (ZipInputStream zis = new ZipInputStream(new FileInputStream(source))) {

        ZipEntry entry = zis.getNextEntry();

        while (entry != null) {

            File file = new File(out, entry.getName());

            if (entry.isDirectory()) {
                file.mkdirs();
            } else {
                File parent = file.getParentFile();

                if (!parent.exists()) {
                    parent.mkdirs();
                }

                try (BufferedOutputStream bos = new BufferedOutputStream(new FileOutputStream(file))) {

                    byte[] buffer = new byte[Math.toIntExact(entry.getSize())];

                    int location;

                    while ((location = zis.read(buffer)) != -1) {
                        bos.write(buffer, 0, location);
                    }
                }
            }
            entry = zis.getNextEntry();
        }
    }
}

Answer 5

最简洁，无库的Java 7+变体：

public static void unzip(InputStream is, Path targetDir) throws IOException {
    try (ZipInputStream zipIn = new ZipInputStream(is)) {
        for (ZipEntry ze; (ze = zipIn.getNextEntry()) != null; ) {
            Path resolvedPath = targetDir.resolve(ze.getName());
            if (ze.isDirectory()) {
                Files.createDirectories(resolvedPath);
            } else {
                Files.createDirectories(resolvedPath.getParent());
                Files.copy(zipIn, resolvedPath);
            }
        }
    }
}

Answer 6

这里有更多＆＃34;现代＆＃34;完整代码基于this帖子但已重构（并使用Lombok）：

import lombok.var;
import lombok.val;

import java.io.File;
import java.io.FileInputStream;
import java.io.FileOutputStream;
import java.io.IOException;
import java.util.zip.ZipInputStream;

import static java.nio.file.Files.createDirectories;

public class UnZip
{
    public static void unZip(String sourceZipFile, String outputDirectory) throws IOException
    {
        val folder = new File(outputDirectory);
        createDirectories(folder.toPath());

        try (val zipInputStream = new ZipInputStream(new FileInputStream(sourceZipFile, Charset.forName("Cp437"))))
        {
            var nextEntry = zipInputStream.getNextEntry();

            while (nextEntry != null)
            {
                val fileName = nextEntry.getName();
                val newFile = new File(outputDirectory + File.separator + fileName);

                newFile.getParentFile().mkdirs();

                if(fileName.endsWith("/")){
                    newFile.mkdirs();
                } else {
                    writeFile(zipInputStream, newFile);
                }

                writeFile(zipInputStream, newFile);

                nextEntry = zipInputStream.getNextEntry();
            }

            zipInputStream.closeEntry();
        }
    }

    private static void writeFile(ZipInputStream inputStream, File file) throws IOException
    {
        val buffer = new byte[1024];
        file.createNewFile();
        try (val fileOutputStream = new FileOutputStream(file))
        {
            int length;
            while ((length = inputStream.read(buffer)) > 0)
            {
                fileOutputStream.write(buffer, 0, length);
            }
        }
    }
}

Answer 7

这是我用来解压缩具有多个目录的zip文件的代码。没有使用外部库。

import java.io.BufferedInputStream;
import java.io.File;
import java.io.FileOutputStream;
import java.io.IOException;
import java.util.Enumeration;
import java.util.zip.ZipEntry;
import java.util.zip.ZipFile;

public class UnzipFile
{
  public static void main(String[] args) throws IOException
  {
    String fileZip = "src/main/resources/abcd/abc.zip";
    File destDir = new File("src/main/resources/abcd/abc");

    try (ZipFile file = new ZipFile(fileZip))
    {
      Enumeration<? extends ZipEntry> zipEntries = file.entries();
      while (zipEntries.hasMoreElements())
      {
        ZipEntry zipEntry = zipEntries.nextElement();
        File newFile = new File(destDir, zipEntry.getName());

        //create sub directories
        newFile.getParentFile().mkdirs();

        if (!zipEntry.isDirectory())
        {
          try (FileOutputStream outputStream = new FileOutputStream(newFile))
          {
            BufferedInputStream inputStream = new BufferedInputStream(file.getInputStream(zipEntry));
            while (inputStream.available() > 0)
            {
              outputStream.write(inputStream.read());
            }
            inputStream.close();
          }
        }

      }
    }
  }

}

Answer 8

您应该从zip文件中获取所有条目：

Enumeration entries = zipFile.getEntries();

然后itareting这个枚举从它获取ZipEntry，检查它是否是一个目录，并创建dirctrory或只是分别提取文件。

Java ZIP - 如何解压缩文件夹？

8 个答案: