我应该如何提取java中的压缩文件夹？

Question

我正在使用以下代码在 Java 中提取 zip 文件。

import java.io.*;
import java.util.zip.*;

class  testZipFiles 
{
    public static void main(String[] args) 
    {

        try
        {
            String filename = "C:\\zip\\includes.zip";
            testZipFiles list = new testZipFiles( );
            list.getZipFiles(filename);
        }
        catch (Exception e)
        {
            e.printStackTrace();
        }
    }

    public void getZipFiles(String filename)
    {
        try
        {
            String destinationname = "c:\\zip\\";
            byte[] buf = new byte[1024];
            ZipInputStream zipinputstream = null;
            ZipEntry zipentry;
            zipinputstream = new ZipInputStream(
            new FileInputStream(filename));

            zipentry = zipinputstream.getNextEntry();
            while (zipentry != null) 
            { 
                //for each entry to be extracted
                String entryName = zipentry.getName();
                System.out.println("entryname "+entryName);
                int n;
                FileOutputStream fileoutputstream;
                File newFile = new File(entryName);
                String directory = newFile.getParent();

                if(directory == null)
                {
                    if(newFile.isDirectory())
                    break;
                }

                fileoutputstream = new FileOutputStream(
                   destinationname+entryName);             

                while ((n = zipinputstream.read(buf, 0, 1024)) > -1)
                    fileoutputstream.write(buf, 0, n);

                fileoutputstream.close(); 
                zipinputstream.closeEntry();
                zipentry = zipinputstream.getNextEntry();

            }//while

            zipinputstream.close();
        }
        catch (Exception e)
       {
            e.printStackTrace();
       }
  }

}

显然，由于break语句，这不会提取文件夹树。 我尝试使用递归来处理文件夹树，但失败了。 有人可以告诉我如何改进此代码以处理文件夹树而不是压缩的单级文件夹。

Answer 1

您可以使用 File.mkdirs() 创建文件夹。 尝试像这样改变你的方法：

public static void getZipFiles(String filename) {
    try {
        String destinationname = "c:\\zip\\";
        byte[] buf = new byte[1024];
        ZipInputStream zipinputstream = null;
        ZipEntry zipentry;
        zipinputstream = new ZipInputStream(
                new FileInputStream(filename));

        zipentry = zipinputstream.getNextEntry();
        while (zipentry != null) {
            //for each entry to be extracted
            String entryName = destinationname + zipentry.getName();
            entryName = entryName.replace('/', File.separatorChar);
            entryName = entryName.replace('\\', File.separatorChar);
            System.out.println("entryname " + entryName);
            int n;
            FileOutputStream fileoutputstream;
            File newFile = new File(entryName);
            if (zipentry.isDirectory()) {
                if (!newFile.mkdirs()) {
                    break;
                }
                zipentry = zipinputstream.getNextEntry();
                continue;
            }

            fileoutputstream = new FileOutputStream(entryName);

            while ((n = zipinputstream.read(buf, 0, 1024)) > -1) {
                fileoutputstream.write(buf, 0, n);
            }

            fileoutputstream.close();
            zipinputstream.closeEntry();
            zipentry = zipinputstream.getNextEntry();

        }//while

        zipinputstream.close();
    } catch (Exception e) {
        e.printStackTrace();
    }
}

Answer 2

另一个选项是 commons-compress，上面链接的网站上有示例代码。

Answer 3

我需要这样做是因为我使用的 API 需要一个 File 参数，而您无法从 JAR 中的资源中获取该参数。

我发现@Emre 的答案不能正常工作。 由于某种原因，ZipEntry 跳过了 JAR 中的一些文件（没有明显的模式）。 我通过使用 JarEntry 解决了这个问题。 上面的代码中还有一个错误，即可以在目录之前枚举 zip 条目中的文件，这会导致异常，因为尚未创建目录。

请注意，以下代码依赖于 Apache Commons 实用程序类。

/**
 * 
 * Extract a directory in a JAR on the classpath to an output folder.
 * 
 * Note: User's responsibility to ensure that the files are actually in a JAR.
 * The way that I do this is to get the URI with
 *     URI url = getClass().getResource("/myresource").toURI();
 * and then if url.isOpaque() we are in a JAR. There may be a more reliable
 * way however, please edit this answer if you know of one.
 * 
 * @param classInJar A class in the JAR file which is on the classpath
 * @param resourceDirectory Path to resource directory in JAR
 * @param outputDirectory Directory to write to  
 * @return String containing the path to the folder in the outputDirectory
 * @throws IOException
 */
private static String extractDirectoryFromClasspathJAR(Class<?> classInJar, String resourceDirectory, String outputDirectory)
        throws IOException {

    resourceDirectory = StringUtils.strip(resourceDirectory, "\\/") + File.separator;

    URL jar = classInJar.getProtectionDomain().getCodeSource().getLocation();
    //Note: If you want to extract from a named JAR, remove the above 
    //line and replace "jar.getFile()" below with the path to the JAR.
    JarFile jarFile = new JarFile(new File(jar.getFile()));

    byte[] buf = new byte[1024];
    Enumeration<JarEntry> jarEntries = jarFile.entries();
    while (jarEntries.hasMoreElements()) {
        JarEntry jarEntry = jarEntries.nextElement();

        if (jarEntry.isDirectory() || !jarEntry.getName().startsWith(resourceDirectory)) {
            continue;               
        }


        String outputFileName = FilenameUtils.concat(outputDirectory, jarEntry.getName());
        //Create directories if they don't exist
        new File(FilenameUtils.getFullPath(outputFileName)).mkdirs();

        //Write file
        FileOutputStream fileOutputStream = new FileOutputStream(outputFileName);
        int n;
        InputStream is = jarFile.getInputStream(jarEntry);
        while ((n = is.read(buf, 0, 1024)) > -1) {
            fileOutputStream.write(buf, 0, n);
        }
        is.close();
        fileOutputStream.close();
    }
    jarFile.close();

    String fullPath = FilenameUtils.concat(outputDirectory, resourceDirectory);
    return fullPath;
}