CODEWITHSUNDEEP
python
David542
David542

Reputation: 110093

Python way to urlquote "non-standard" parts of url

I have the following url, that I have:

https://www.verizon.com/OnDemand/TVShows/TVShowDetails/Sr. Avila/1/9

I would like to encode it so that it looks like a normal url, but is valid. For example:

https://www.verizon.com/OnDemand/TVShows/TVShowDetails/Sr.%20Avila/1/9

However, if I use the standard urllib.quote it encodes everything:

>>> urllib.quote('https://www.verizon.com/OnDemand/TVShows/TVShowDetails/Sr. Avila/1/9')
'https%3A//www.verizon.com/OnDemand/TVShows/TVShowDetails/Sr.%20Avila/1/9'

Is there a python method that will encode only the non-standard parts of the url, i.e., excluding the forward slashes and colons, etc?

Upvotes: 2

Views: 376

Answers (2)

Kelvin
Kelvin

Reputation: 1367

An example, for Python2

In [45]: scheme, netloc, path, query, fragment = urllib2.urlparse.urlsplit(url)
In [60]: urllib2.urlparse.urlunsplit([scheme, netloc, urllib.quote(path), query, fragment])
Out[60]: 'https://www.verizon.com/OnDemand/TVShows/TVShowDetails/Sr.%20Avila/1/9'

Upvotes: 1

jeremycg
jeremycg

Reputation: 24945

You want the 'safe' argument:

If you are on Python3, using urllib.parse:

import urllib.parse

x ='https://www.verizon.com/OnDemand/TVShows/TVShowDetails/Sr. Avila/1/9'
urllib.parse.quote(x, safe = ':/')

out:

 'https://www.verizon.com/OnDemand/TVShows/TVShowDetails/Sr.%20Avila/1/9'

Upvotes: 2

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