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Unzip Files Effortlessly with Go
Unzipping files in Go should be a breeze, but sometimes you encounter obstacles. Let's explore a straightforward solution to unzip files effectively.
Initial Code:
You presented the following code snippet:
func Unzip(src, dest string) error { r, err := zip.OpenReader(src) if err != nil { return err } defer r.Close() for _, f := range r.File { rc, err := f.Open() if err != nil { return err } defer rc.Close() path := filepath.Join(dest, f.Name) if f.FileInfo().IsDir() { os.MkdirAll(path, f.Mode()) } else { f, err := os.OpenFile( path, os.O_WRONLY|os.O_CREATE|os.O_TRUNC, f.Mode()) if err != nil { return err } defer f.Close() _, err = io.Copy(f, rc) if err != nil { return err } } } return nil }
Optimized Solution:
To improve upon the initial code, consider the following enhancements:
func Unzip(src, dest string) error { r, err := zip.OpenReader(src) if err != nil { return err } defer func() { if err := r.Close(); err != nil { panic(err) } }() os.MkdirAll(dest, 0755) // Closure to address file descriptors issue extractAndWriteFile := func(f *zip.File) error { rc, err := f.Open() if err != nil { return err } defer func() { if err := rc.Close(); err != nil { panic(err) } }() path := filepath.Join(dest, f.Name) // Check for ZipSlip (Directory traversal) if !strings.HasPrefix(path, filepath.Clean(dest) + string(os.PathSeparator)) { return fmt.Errorf("illegal file path: %s", path) } if f.FileInfo().IsDir() { os.MkdirAll(path, f.Mode()) } else { os.MkdirAll(filepath.Dir(path), f.Mode()) f, err := os.OpenFile(path, os.O_WRONLY|os.O_CREATE|os.O_TRUNC, f.Mode()) if err != nil { return err } defer func() { if err := f.Close(); err != nil { panic(err) } }() _, err = io.Copy(f, rc) if err != nil { return err } } return nil } for _, f := range r.File { err := extractAndWriteFile(f) if err != nil { return err } } return nil }
Improvements:
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