如何从目录路径列表中打开文件？

import os

def open_file_from_directory(directory_list, file_name):
    for directory in directory_list:
        file_path = os.path.join(directory, file_name)
        if os.path.isfile(file_path):
            with open(file_path, 'r') as file:
                # 在这里可以进行文件的读取操作
                file_content = file.read()
                print(file_content)
            break
    else:
        print("文件不存在")

# 示例调用
directory_list = ["/path/to/directory1", "/path/to/directory2"]
file_name = "example.txt"
open_file_from_directory(directory_list, file_name)