
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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