如何压缩和解压文件？

Question

如何对已经在 DDMS 中的文件进行压缩和解压缩：data/data/mypackage/files/ 我需要一个简单的例子。我已经搜索过有关压缩和解压缩的相关内容，但是没有任何一个例子适合我。有人可以提供一些例子吗？先谢谢了。

- user905216

Answer 1

查看java.util.zip.* 类以获取zip功能。我已经编写了一些基本的zip / unzip代码，如下所示。希望对您有所帮助。

public static void zip(String[] files, String zipFile) throws IOException {
    BufferedInputStream origin = null;
    ZipOutputStream out = new ZipOutputStream(new BufferedOutputStream(new FileOutputStream(zipFile)));
    try { 
        byte data[] = new byte[BUFFER_SIZE];

        for (int i = 0; i < files.length; i++) {
            FileInputStream fi = new FileInputStream(files[i]);    
            origin = new BufferedInputStream(fi, BUFFER_SIZE);
            try {
                ZipEntry entry = new ZipEntry(files[i].substring(files[i].lastIndexOf("/") + 1));
                out.putNextEntry(entry);
                int count;
                while ((count = origin.read(data, 0, BUFFER_SIZE)) != -1) {
                    out.write(data, 0, count);
                }
            }
            finally {
                origin.close();
            }
        }
    }
    finally {
        out.close();
    }
}

public static void unzip(String zipFile, String location) throws IOException {
    try {
        File f = new File(location);
        if(!f.isDirectory()) {
            f.mkdirs();
        }
        ZipInputStream zin = new ZipInputStream(new FileInputStream(zipFile));
        try {
            ZipEntry ze = null;
            while ((ze = zin.getNextEntry()) != null) {
                String path = location + ze.getName();

                if (ze.isDirectory()) {
                    File unzipFile = new File(path);
                    if(!unzipFile.isDirectory()) {
                        unzipFile.mkdirs();
                    }
                }
                else {
                    FileOutputStream fout = new FileOutputStream(path, false);
                    try {
                        for (int c = zin.read(); c != -1; c = zin.read()) {
                            fout.write(c);
                        }
                        zin.closeEntry();
                    }
                    finally {
                        fout.close();
                    }
                }
            }
        }
        finally {
            zin.close();
        }
    }
    catch (Exception e) {
        Log.e(TAG, "Unzip exception", e);
    }
}

- brianestey

---

这是错误的：String path = location + ze.getName(); if (ze.isDirectory()) { File unzipFile = new File(path); if(!unzipFile.isDirectory()) { unzipFile.mkdirs(); } } - maohieng

缓冲区大小变量应该是多少？ - user6649667

这在Android 5.1上运行得非常好。谢谢！我在这里添加了一个答案，它是你的修改版本，接受List<File> files和File zipFile作为参数，可能对其他人有帮助。 - Joshua Pinter

Answer 2

brianestey提供的zip函数很好用，但是unzip函数由于每次只读取一个字节而非常慢。这里是他的unzip函数的改进版本，利用缓冲区更快。

/**
 * Unzip a zip file.  Will overwrite existing files.
 * 
 * @param zipFile Full path of the zip file you'd like to unzip.
 * @param location Full path of the directory you'd like to unzip to (will be created if it doesn't exist).
 * @throws IOException
 */
public static void unzip(String zipFile, String location) throws IOException {
    int size;
    byte[] buffer = new byte[BUFFER_SIZE];

    try {
        if ( !location.endsWith(File.separator) ) {
            location += File.separator;
        }
        File f = new File(location);
        if(!f.isDirectory()) {
            f.mkdirs();
        }
        ZipInputStream zin = new ZipInputStream(new BufferedInputStream(new FileInputStream(zipFile), BUFFER_SIZE));
        try {
            ZipEntry ze = null;
            while ((ze = zin.getNextEntry()) != null) {
                String path = location + ze.getName();
                File unzipFile = new File(path);

                if (ze.isDirectory()) {
                    if(!unzipFile.isDirectory()) {
                        unzipFile.mkdirs();
                    }
                } else {
                    // check for and create parent directories if they don't exist
                    File parentDir = unzipFile.getParentFile();
                    if ( null != parentDir ) {
                        if ( !parentDir.isDirectory() ) {
                            parentDir.mkdirs();
                        }
                    }

                    // unzip the file
                    FileOutputStream out = new FileOutputStream(unzipFile, false);
                    BufferedOutputStream fout = new BufferedOutputStream(out, BUFFER_SIZE);
                    try {
                        while ( (size = zin.read(buffer, 0, BUFFER_SIZE)) != -1 ) {
                            fout.write(buffer, 0, size);
                        }

                        zin.closeEntry();
                    }
                    finally {
                        fout.flush();
                        fout.close();
                    }
                }
            }
        }
        finally {
            zin.close();
        }
    }
    catch (Exception e) {
        Log.e(TAG, "Unzip exception", e);
    }
}

- Ben

---

它是否真的会“抛出IOException”？每个异常都会被“catch(Exception e)”捕获。 - dragi

非常感谢你，你节省了我的时间。当ze.getName()并非直接文件名，而可能包含文件夹和该文件夹内的文件时，这也非常有用。例如：1/abc.txt。再次感谢你。 - Hiren Dabhi

我使用了BUFFER_SIZE作为8192。它运行得非常好。 - chisom emmanuel

Answer 3

本答案基于brianestey、Ben、Giacomo Mattiuzzi和Joshua Pinter的回答。

将函数重写为Kotlin，并添加用于处理Uri的函数。

import android.content.Context
import android.net.Uri
import java.io.*
import java.util.zip.ZipEntry
import java.util.zip.ZipInputStream
import java.util.zip.ZipOutputStream

private const val MODE_WRITE = "w"
private const val MODE_READ = "r"

fun zip(zipFile: File, files: List<File>) {
    ZipOutputStream(BufferedOutputStream(FileOutputStream(zipFile))).use { outStream ->
        zip(outStream, files)
    }
}

fun zip(context: Context, zipFile: Uri, files: List<File>) {
    context.contentResolver.openFileDescriptor(zipFile, MODE_WRITE).use { descriptor ->
        descriptor?.fileDescriptor?.let {
            ZipOutputStream(BufferedOutputStream(FileOutputStream(it))).use { outStream ->
                zip(outStream, files)
            }
        }
    }
}

private fun zip(outStream: ZipOutputStream, files: List<File>) {
    files.forEach { file ->
        outStream.putNextEntry(ZipEntry(file.name))
        BufferedInputStream(FileInputStream(file)).use { inStream ->
            inStream.copyTo(outStream)
        }
    }
}

fun unzip(zipFile: File, location: File) {
    ZipInputStream(BufferedInputStream(FileInputStream(zipFile))).use { inStream ->
        unzip(inStream, location)
    }
}

fun unzip(context: Context, zipFile: Uri, location: File) {
    context.contentResolver.openFileDescriptor(zipFile, MODE_READ).use { descriptor ->
        descriptor?.fileDescriptor?.let {
            ZipInputStream(BufferedInputStream(FileInputStream(it))).use { inStream ->
                unzip(inStream, location)
            }
        }
    }
}

private fun unzip(inStream: ZipInputStream, location: File) {
    if (location.exists() && !location.isDirectory)
        throw IllegalStateException("Location file must be directory or not exist")

    if (!location.isDirectory) location.mkdirs()

    val locationPath = location.absolutePath.let {
        if (!it.endsWith(File.separator)) "$it${File.separator}"
        else it
    }

    var zipEntry: ZipEntry?
    var unzipFile: File
    var unzipParentDir: File?

    while (inStream.nextEntry.also { zipEntry = it } != null) {
        unzipFile = File(locationPath + zipEntry!!.name)
        if (zipEntry!!.isDirectory) {
            if (!unzipFile.isDirectory) unzipFile.mkdirs()
        } else {
            unzipParentDir = unzipFile.parentFile
            if (unzipParentDir != null && !unzipParentDir.isDirectory) {
                unzipParentDir.mkdirs()
            }
            BufferedOutputStream(FileOutputStream(unzipFile)).use { outStream ->
                inStream.copyTo(outStream)
            }
        }
    }
}

- Pavel Shirokov

Answer 4

使用File代替文件路径String

本答案基于@brianestey的优秀答案。

我修改了他的zip方法，使其接受File列表而不是文件路径，并接受输出zip文件的File而不是文件路径。如果用户正在处理这些内容，这可能会对其他人有所帮助。

public static void zip( List<File> files, File zipFile ) throws IOException {
    final int BUFFER_SIZE = 2048;

    BufferedInputStream origin = null;
    ZipOutputStream out = new ZipOutputStream(new BufferedOutputStream(new FileOutputStream(zipFile)));

    try {
        byte data[] = new byte[BUFFER_SIZE];

        for ( File file : files ) {
            FileInputStream fileInputStream = new FileInputStream( file );

            origin = new BufferedInputStream(fileInputStream, BUFFER_SIZE);

            String filePath = file.getAbsolutePath();

            try {
                ZipEntry entry = new ZipEntry( filePath.substring( filePath.lastIndexOf("/") + 1 ) );

                out.putNextEntry(entry);

                int count;
                while ((count = origin.read(data, 0, BUFFER_SIZE)) != -1) {
                    out.write(data, 0, count);
                }
            }
            finally {
                origin.close();
            }
        }
    }
    finally {
        out.close();
    }
}

- Joshua Pinter