如何从内容字符串中获取文件路径和文件名？

import re

def get_file_path_and_name(content):
    pattern = r'(.*/)(.*)'
    match = re.match(pattern, content)
    if match:
        file_path = match.group(1)
        file_name = match.group(2)
        return file_path, file_name
    else:
        return None

# 示例用法
content = '/path/to/file.txt'
result = get_file_path_and_name(content)
if result:
    file_path, file_name = result
    print("文件路径：", file_path)
    print("文件名：", file_name)
else:
    print("无法提取文件路径和文件名")