How to extract files from a zip archive in Go?

To extract files from a zip archive in Go, you can make use of the archive/zip package provided by the standard library. Here's an example of how you can do it:

package main

import (
	"archive/zip"
	"fmt"
	"io"
	"log"
	"os"
	"path/filepath"
)

func main() {
	// Open the zip file
	zipFile, err := zip.OpenReader("path/to/archive.zip")
	if err != nil {
		log.Fatal(err)
	}
	defer zipFile.Close()

	// Create a directory for extracted files
	extractDir := "path/to/extract"
	os.MkdirAll(extractDir, os.ModePerm)

	// Extract each file from the zip archive
	for _, file := range zipFile.File {
		destPath := filepath.Join(extractDir, file.Name)

		// If the file is a directory, create it
		if file.FileInfo().IsDir() {
			os.MkdirAll(destPath, os.ModePerm)
			continue
		}

		// Open the source file inside the zip
		srcFile, err := file.Open()
		if err != nil {
			log.Fatal(err)
		}
		defer srcFile.Close()

		// Create the destination file
		destFile, err := os.OpenFile(
			destPath,
			os.O_WRONLY|os.O_CREATE|os.O_TRUNC,
			file.Mode(),
		)
		if err != nil {
			log.Fatal(err)
		}
		defer destFile.Close()

		// Copy the contents of the source file to destination file
		_, err = io.Copy(destFile, srcFile)
		if err != nil {
			log.Fatal(err)
		}

		// Optionally, you can print the extracted file name
		fmt.Println("Extracted:", file.Name)
	}

	fmt.Println("Extraction completed.")
}

Make sure to replace "path/to/archive.zip" with the actual path to your zip archive file. Also, change "path/to/extract" to the desired directory where you want to extract the files.

This code will open the zip file, create a directory for the extracted files, and then extract each file from the archive one by one. If a file inside the archive has a directory structure, it will be created before extracting the file. Finally, it will print the names of the extracted files.