def generate_dockerfile_command()

in scripts/gen_dockerfile.py [0:0]

• 10 lines of code
• 3 McCabe index (conditional complexity)

def generate_dockerfile_command(base_image, config_file, source_dir):
    """Write a Dockerfile and helper files for an application.

    Args:
        base_image (str): Docker image name to build on top of
        config_file (str): Path to user's app.yaml (might be <service>.yaml)
        source_dir (str): Directory container user's source code
    """
    # Read yaml file.  Does not currently support multiple services
    # with configuration filenames besides app.yaml
    with io.open(config_file, 'r', encoding='utf8') as yaml_config_file:
        raw_config = yaml.safe_load(yaml_config_file)

    # Determine complete configuration
    app_config = get_app_config(raw_config, base_image, config_file,
                                  source_dir)

    # Generate list of filenames and their textual contents
    files = generate_files(app_config)

    # Write files
    for filename, contents in files.items():
        full_filename = os.path.join(source_dir, filename)
        with io.open(full_filename, 'w', encoding='utf8') as outfile:
            outfile.write(contents)