CODEWITHSUNDEEP
python-3.x
Duke Dougal
Duke Dougal

Reputation: 26406

How to use urllib.request.DataHandler in Python 3.6?

I want to open a datauri in Python 3.6

It is described here: https://docs.python.org/3/library/urllib.request.html

But there's really no practical example of how to use it.

I've tried the following without luck. Can anyone suggest the right way to get the data from a datauri using urllib.request.DataHandler?

>>> req = urllib.request.Request("data:text/html,%3Ch1%3EHello%2C%20World!%3C%2Fh1%3E")
>>> with urllib.request.DataHandler.data_open(req) as response:
... data = response.read()
  File "<stdin>", line 2
    data = response.read()
       ^
IndentationError: expected an indented block
>>>     data = response.read()
  File "<stdin>", line 1
    data = response.read()
    ^
IndentationError: unexpected indent
>>> with urllib.request.DataHandler.data_open(req) as response:
...     data = response.read()
...
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: data_open() missing 1 required positional argument: 'req'
>>> with urllib.request.DataHandler.data_open(req) as response:
...     data = response.read()
...
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: data_open() missing 1 required positional argument: 'req'
>>>

Upvotes: 1

Views: 1509

Answers (1)

jfs
jfs

Reputation: 414695

You could pass data URI to urlopen():

#!/usr/bin/env python3
from urllib.request import urlopen

url = "data:text/html,%3Ch1%3EHello%2C%20World!%3C%2Fh1%3E"
with urlopen(url) as response:
    data = response.read()

print(data.decode())

Upvotes: 3

Related Questions