Why does urlunparse return extra slash after manipulating scheme?

Question

I need to add a scheme to urls which don't have it. I want to use the following code:

from urllib.parse import urlparse, urlunparse

url = 'example.com'
parsed = urlparse(url)
parsed = parsed._replace(scheme='https')
new_url = urlunparse(parsed)
print(new_url)

Instead of this:

https://example.com

the script is returning this:

https:///example.com

which throws and error if I try to get the url like so:

requests.get('https:///example.com')

Why is this happening and what can I do about it?

I am using:

Windows 10
Python 3.6.1
Anaconda 4.4.0

Three slashes is due to your urls is considered as path. – shaik moeed May 21 '19 at 16:38 — shaik moeed, May 21 '19 at 16:38

score 1 · Accepted Answer · answered May 21 '19 at 16:37

1

The initial string is parsed as a path component because there is no scheme to indicate that the string is a host:

urlparse('example.com')
# ParseResult(scheme='', netloc='', path='example.com', params='', query='', fragment='')

Perhaps add a scheme to make this explicit:

urlparse('http://example.com')
# ParseResult(scheme='http', netloc='example.com', path='', params='', query='', fragment='')

answered May 21 '19 at 16:37

jspcal

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This is duplicate question of [this](https://stackoverflow.com/a/7289533/8353711). – shaik moeed May 21 '19 at 16:41

Why does urlunparse return extra slash after manipulating scheme?

1 Answers1