如何在 python 中创建受密码保护的 csv 或 ZIP 存档

Question

async def get_report(id):
    try:
        data = await collection.find({"_id": ObjectId(id)}).to_list(None)
        if data:
            df1 = pd.DataFrame(data)
            directory = Path("/")
            file_name = "abcd.csv"
            file_path = directory / file_name
            df1.to_csv(file_path, index=False)
            password = "1234"
            zip_file_path = directory / (file_name + ".zip")
            with zipfile.ZipFile(zip_file_path, 'w', zipfile.ZIP_DEFLATED) as zf:
                zf.setpassword(password.encode('utf-8'))
                zf.write(file_path, arcname=file_name)
            response = FileResponse(zip_file_path, media_type='application/zip', filename=file_name+".zip")
            return response
            #json_data = {'file':str(zip_file_path)}
            #response = requests.post(url,json=json_data)
    except Exception as e:
        print(e)

这是我的代码，在这里我尝试创建数据框的 csv 并在电子邮件中发送文件（代码不存在，但 csv 正在邮件中发送（其工作））文件正在下载 csv 和 zip 两者。但这里没有密码保护（返回响应），是否可以做，如果是的话，我在这里缺少什么。

Answer 1

内置的zipfile模块不支持写入密码加密文件（仅读取）。你可以使用 pyminizip:

pip install pyminizip

import pyminizip


async def get_report(id):
    try:
        data = await collection.find({"_id": ObjectId(id)}).to_list(None)
        if data:
            df1 = pd.DataFrame(data)
            directory = Path("/")
            file_name = "abcd.csv"
            file_path = directory / file_name
            df1.to_csv(file_path, index=False)# remove if you dont want to save .csv file as well
            password = "1234"
            zip_file_path = directory / (file_name. + ".zip")
            pyminizip.compress(file_path, None, zip_file_path, password, 0)
            response = FileResponse(zip_file_path, media_type='application/zip', filename=file_name+".zip")
            return response
            #json_data = {'file':str(zip_file_path)}
            #response = requests.post(url,json=json_data)
    except Exception as e:
        print(e)