I’m using BytesIO on some of my python scripts. Writing a file into memory to upload it to SharePoint. I think I don’t understand how that works quite enough or what a buffer is.

My concern (possibly unwarranted) is that I need to close the function/buffer once the files been successfully uploaded. Or is that unnecessary and the buffer gets flushed automatically

Hello all,

I'm trying to wrap my head around the practical differences between:

<class 'bytes'> and <class '_io.BytesIO'>.

I read through the documentation:

https://docs.python.org/3/library/io.html?highlight=bytesio#binary-i-o

Binary I/O (also called buffered I/O) expects bytes-like objects and produces bytes objects. No encoding, decoding, or newline translation is performed. This category of streams can be used for all kinds of non-text data, and also when manual control over the handling of text data is desired.

It provides some examples:

The easiest way to create a binary stream is with open() with 'b' in the mode string:

and

f = io.BytesIO(b"some initial binary data: \x00\x01")

So I read all this, but so what? Why would you use the io.BytesIO data type over a standard bytes data type?

EDIT: Let me provide some additional context that I just discovered after reading the documentation on lxml.

https://lxml.de/parsing.html#parsing-html

I'm using the requests object and parsing the results with lxml. Here is the example code:

from io import BytesIO
from lxml import etree
#* etree - https://lxml.de/parsing.html
#? etree stands for element tree

import requests

#? Need to know concepts
#?  What are bytes
#?  HTTP status codes
#?  HTTP methods (GET. POST, PUT, DELETE)
#?  bytes - https://docs.python.org/3/library/stdtypes.html?highlight=bytes#bytes-objects

url = 'http://localhost'
#! The URL https://nostarch.com/ doesn't seem to work

resp = requests.get(url=url)
html_bytes = resp.content
parser = etree.HTMLParser()
content = etree.parse(BytesIO(html_bytes), parser=parser)

print(type(html_bytes))
print(type(BytesIO(html_bytes)))

for link in content.findall('//a'):
    print(f"{link.get('href')} -> {link.text}")

Kind regards