Wie kann ich Dateien in Go unter angemessener Fehlerbehandlung und Sicherheitsaspekten effizient entpacken?

Unzip Files Effortlessly with Go

Unzipping files in Go is a straightforward task with the right tools. The zip package in Go provides a convenient way to extract files from ZIP archives.

Current Code

The code snippet provided initializes a zip reader, iterates over the files in the archive, and extracts them to the designated destination. The nested defer statements can lead to issues, as noted by @Nick Craig-Wood.

Improved Solution

To address this issue, a closure is introduced to encapsulate the file extraction and writing logic. Additionally, error handling is added to the Close() calls for both the zip reader and the individual file readers:

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)

    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)
            }
        }()

        ... (Code for file extraction and writing goes here) ...

        return nil
    }

    for _, f := range r.File {
        err := extractAndWriteFile(f)
        if err != nil {
            return err
        }
    }

    return nil
}

This improved solution creates the destination directory if it doesn't exist and ensures proper error handling for all file descriptors and closure cleanup.

Additional Considerations

The updated code also includes ZipSlip detection to prevent directory traversal and potential security risks associated with extracting files outside of the designated destination path.

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