Spring.breakers.2012.480p.vegamovies.nl.mkv — !!better!!

# Usage file_name = "Spring.Breakers.2012.480p.Vegamovies.NL.mkv" movie_info = extract_movie_info(file_name) print_movie_info(movie_info) This example demonstrates a simple way to extract and display information from a movie file name. Depending on the specific requirements, you might need to adjust the regular expressions used for parsing the file name.

def print_movie_info(info): print("Movie File Information:") for key, value in info.items(): if key == 'file_name': print(f"- **File Name:** {value}") elif key == 'title': print(f"- **Movie Title:** {value}") elif key == 'year': print(f"- **Release Year:** {value}") elif key == 'resolution': print(f"- **Resolution:** {value}") elif key == 'source': print(f"- **Source:** {value}") elif key == 'format': print(f"- **File Format:** {value}") Spring.Breakers.2012.480p.Vegamovies.NL.mkv

def extract_movie_info(file_name): # Assuming a common naming convention patterns = { 'title': r'^([\w\s]+)\.', 'year': r'(\d{4})\.', 'resolution': r'(\d+p)\.', 'source': r'(\w+)\.', 'format': r'\.(\w+)$' } info = {} info['file_name'] = file_name # Extract title match = re.search(patterns['title'], file_name) info['title'] = match.group(1).replace('.','') if match else "Unknown" # Extract year match = re.search(patterns['year'], file_name) info['year'] = match.group(1) if match else "Unknown" # Extract resolution match = re.search(patterns['resolution'], file_name) info['resolution'] = match.group(1) if match else "Unknown" # Extract source match = re.search(patterns['source'], file_name) info['source'] = match.group(1) if match else "Unknown" # Extract file format match = re.search(patterns['format'], file_name) info['format'] = match.group(1) if match else "Unknown" return info # Usage file_name = "Spring

- **File Name:** Spring.Breakers.2012.480p.Vegamovies.NL.mkv - **Movie Title:** Spring Breakers - **Release Year:** 2012 - **Resolution:** 480p - **Source:** Vegamovies - **File Format:** MKV import re file_name) info['title'] = match.group(1).replace('.'