有没有人有一个脚本来计算包含特定单词的连续文件的数量？

import os

def count_files_with_word(directory, word):
    count = 0
    for root, dirs, files in os.walk(directory):
        for file in files:
            file_path = os.path.join(root, file)
            if os.path.isfile(file_path):
                with open(file_path, 'r') as f:
                    content = f.read()
                    if word in content:
                        count += 1
    return count

directory = '/path/to/directory'  # 替换为要搜索的目录路径
word = '特定单词'  # 替换为要搜索的特定单词

result = count_files_with_word(directory, word)
print(f"包含特定单词的连续文件数量：{result}")