为由 urlopen() 或 requests.get() 创建的类文件对象提供文件名 [英] supply a filename for a file-like object created by urlopen() or requests.get()

问题描述

我正在使用 Telepot 库构建 Telegram 机器人.要发送从 Internet 下载的图片，我必须使用 sendPhoto 方法，它接受一个类似文件的对象.

浏览文档我发现了这个建议:

<块引用>

如果类似文件的对象是通过 urlopen() 获取的，您很可能必须提供文件名，因为 Telegram 服务器需要知道文件扩展名.

所以问题是，如果我通过用 requests.get 打开它并像这样用 BytesIO 包装来获取我的类文件对象:

res = requests.get(some_url)tbot.sendPhoto(信使_id，io.BytesIO(res.content))

如何以及在何处提供文件名?

解决方案

您可以提供文件名作为对象的 .name 属性.

使用 open() 打开文件有一个 .name 属性.

<预><代码>>>>local_file = open(file.txt")>>>本地文件<在地址处打开文件'file.txt'，模式'r'>>>>本地文件名'文件.txt'

打开网址的地方没有.这就是文档特别提到这一点的原因.

<预><代码>>>>导入 urllib>>>url_file = urllib.open("http://example.com/index.html")>>>网址文件<addinfourl at 44 which fp = <open file 'nul', mode 'rb' at ADDRESS>>>>>url_file.nameAttributeError: addinfourl 实例没有属性名称"

在你的情况下，你需要创建类似文件的对象，并给它一个 .name 属性:

res = requests.get(some_url)the_file = io.BytesIO(res.content)the_file.name = 'file.image'tbot.sendPhoto(信使_id，文件)

I am building a Telegram bot using a Telepot library. To send a picture downloaded from Internet I have to use sendPhoto method, which accepts a file-like object.

Looking through the docs I found this advice:

If the file-like object is obtained by urlopen(), you most likely have to supply a filename because Telegram servers require to know the file extension.

So the question is, if I get my filelike object by opening it with requests.get and wrapping with BytesIO like so:

res = requests.get(some_url)
tbot.sendPhoto(
    messenger_id,
    io.BytesIO(res.content)
)

how and where do I supply a filename?

解决方案

You would supply the filename as the object's .name attribute.

Opening a file with open() has a .name attribute.

>>> local_file = open("file.txt")
>>> local_file
<open file 'file.txt', mode 'r' at ADDRESS>
>>> local_file.name
'file.txt'

Where opening a url does not. Which is why the documentation specifically mentions this.

>>> import urllib
>>> url_file = urllib.open("http://example.com/index.html")
>>> url_file
<addinfourl at 44 whose fp = <open file 'nul', mode 'rb' at ADDRESS>>
>>> url_file.name
AttributeError: addinfourl instance has no attribute 'name'

In your case, you would need to create the file-like object, and give it a .name attribute:

res = requests.get(some_url)
the_file = io.BytesIO(res.content)
the_file.name = 'file.image'

tbot.sendPhoto(
    messenger_id,
    the_file
)

这篇关于为由 urlopen() 或 requests.get() 创建的类文件对象提供文件名的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！