如何在Java中实现Zip文件的批量处理

在Java中，可以使用java.util.zip包中的类来处理ZIP文件

1. 添加依赖

如果你使用Maven，可以在pom.xml文件中添加以下依赖：

<dependencies>
    <dependency>
        <groupId>org.apache.commons</groupId>
        <artifactId>commons-compress</artifactId>
        <version>1.21</version>
    </dependency>
</dependencies>

2. 读取ZIP文件

import org.apache.commons.compress.archivers.zip.*;
import java.io.*;
import java.nio.file.*;
import java.util.zip.ZipEntry;

public class UnzipUtility {

    public static void main(String[] args) {
        String zipFilePath = "path/to/your/zipfile.zip";
        String destDirectory = "path/to/your/destination/directory";

        try {
            unzip(zipFilePath, destDirectory);
        } catch (IOException e) {
            e.printStackTrace();
        }
    }

    public static void unzip(String zipFilePath, String destDirectory) throws IOException {
        File destDir = new File(destDirectory);
        if (!destDir.exists()) {
            destDir.mkdir();
        }

        ZipInputStream zipIn = new ZipInputStream(new FileInputStream(zipFilePath));
        ZipEntry entry = zipIn.getNextEntry();

        while (entry != null) {
            String filePath = destDirectory + File.separator + entry.getName();
            if (!entry.isDirectory()) {
                extractFile(zipIn, filePath);
            } else {
                File dir = new File(filePath);
                dir.mkdirs();
            }
            zipIn.closeEntry();
            entry = zipIn.getNextEntry();
        }
        zipIn.close();
    }

    private static void extractFile(ZipInputStream zipIn, String filePath) throws IOException {
        try (BufferedOutputStream bos = new BufferedOutputStream(new FileOutputStream(filePath))) {
            byte[] bytesIn = new byte[4096];
            int read = 0;
            while ((read = zipIn.read(bytesIn)) != -1) {
                bos.write(bytesIn, 0, read);
            }
        }
    }
}

3. 打包ZIP文件

import org.apache.commons.compress.archivers.zip.*;
import java.io.*;
import java.nio.file.*;

public class ZipUtility {

    public static void main(String[] args) {
        String sourceDirectory = "path/to/your/source/directory";
        String zipFilePath = "path/to/your/zipfile.zip";

        try {
            createZip(sourceDirectory, zipFilePath);
        } catch (IOException e) {
            e.printStackTrace();
        }
    }

    public static void createZip(String sourceDirectory, String zipFilePath) throws IOException {
        File sourceDir = new File(sourceDirectory);
        if (!sourceDir.isDirectory()) {
            throw new IllegalArgumentException("Source directory is not a valid directory");
        }

        try (ZipOutputStream zipOut = new ZipOutputStream(new FileOutputStream(zipFilePath))) {
            addDirectoryToZip(sourceDir, sourceDir.getName(), zipOut);
        }
    }

    private static void addDirectoryToZip(File directory, String baseName, ZipOutputStream zipOut) throws IOException {
        File[] files = directory.listFiles();
        if (files != null) {
            for (File file : files) {
                addFileToZip(file, baseName + File.separator + file.getName(), zipOut);
            }
        }
    }

    private static void addFileToZip(File file, String fileName, ZipOutputStream zipOut) throws IOException {
        try (BufferedInputStream bis = new BufferedInputStream(new FileInputStream(file))) {
            byte[] bytesIn = new byte[4096];
            int read = 0;
            while ((read = bis.read(bytesIn)) != -1) {
                zipOut.write(bytesIn, 0, read);
            }
        }
    }
}

这些示例展示了如何读取ZIP文件中的内容并将其提取到目标目录，以及如何将多个文件和目录打包到一个ZIP文件中。你可以根据需要修改这些示例以满足你的需求。