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Created on 2022-01-16 09:44 by scratch, last changed 2022-04-11 14:59 by admin.
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| File name | Uploaded | Description | Edit | |
| urllib_issue.py | scratch, 2022-01-16 09:44 | python version 3.9.7 anaconda | ||
| Messages (2) | |||
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| msg410687 - (view) | Author: Anh Dang (scratch) | Date: 2022-01-16 09:44 | |
urllib.parse.urlencode() return "%2F%3F" instead of "/?" |
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| msg411197 - (view) | Author: Terry J. Reedy (terry.reedy) *  |
Date: 2022-01-21 22:09 | |
'urlencode()' is a TypeError as a query (dict) is needed. The claim is that '/?' in a key are encoded but should not be. I verified the encoding in 3.10.
>>> urlencode({'/?link': 'pubmed'})
'%2F%3Flink=pubmed'
https://docs.python.org/3/library/urllib.parse.html#urllib.parse.urlencode
https://docs.python.org/3/library/urllib.parse.html#urllib.parse.quote
and the following entry from 'quote_plus' say that by default, quote_via is quote_plus and the latter quotes '/' and '?'. So the bug report as stated is not valid.
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They also say that passing 'quote_via=quote' should suppress quoting of '/', because it defaults to 'safe='/', but it does not.
>>> urlencode({'/?link': 'pubmed'}, quote_via=quote)
'%2F%3Flink=pubmed'
So either the doc should be changed in 3 places, or the default safe for quote should be '/' as documented.
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Anh, use safe='/?' to get what you want.
>>> urlencode({'/?link': 'pubmed'}, quote_via=quote, safe='/?')
'/?link=pubmed'
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| History | |||
|---|---|---|---|
| Date | User | Action | Args |
| 2022-04-11 14:59:54 | admin | set | github: 90555 |
| 2022-01-21 22:09:07 | terry.reedy | set | nosy:
+ terry.reedy, orsenthil messages: + msg411197 title: urllib.parse.urlencode() return wrong character -> urllib.parse.quote uses safe='' as default |
| 2022-01-16 09:44:55 | scratch | create | |