Example:
http://example.com/?a=text&q2=text2&q3=text3&q2=text4
After removing "q2", it will return:
http://example.com/?q=text&q3=text3
In this case, there were multiple "q2" and all have been removed.
Add a comment
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9 Answers
import sys
if sys.version_info.major == 3:
from urllib.parse import urlencode, urlparse, urlunparse, parse_qs
else:
from urllib import urlencode
from urlparse import urlparse, urlunparse, parse_qs
url = 'http://example.com/?a=text&q2=text2&q3=text3&q2=text4&b#q2=keep_fragment'
u = urlparse(url)
query = parse_qs(u.query, keep_blank_values=True)
query.pop('q2', None)
u = u._replace(query=urlencode(query, True))
print(urlunparse(u))
Output:
http://example.com/?a=text&q3=text3&b=#q2=keep_fragment
6 Comments
Miki Tebeka
@kurifu The OP wanted to remove only one parameter, not the whole query.
Vanni Totaro
Python 3 imports:
from urllib.parse import urlencode, urlparse, urlunparse, parse_qs
IamMashed
access to a protected member _replace of a class.... how can we avoid this warning message?
Miki Tebeka
@IamMashed This is how namedtuples work - docs.python.org/3/library/… It's probably some kind of linter adding the warning.
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To remove all query string parameters:
from urllib.parse import urljoin, urlparse
url = 'http://example.com/?a=text&q2=text2&q3=text3&q2=text4'
urljoin(url, urlparse(url).path) # 'http://example.com/'
For Python2, replace the import with:
from urlparse import urljoin, urlparse
3 Comments
Dolph
I like this approach a bit better than the popular answer because it doesn't call any internal APIs, but it will also eliminate URL fragments, whereas the popular answer will preserve them. It also doesn't solve the OP's exact question (it deletes all query string parameters), but it solves mine :)
merlin
Was first looking into furl, but this removes the need to install another library. Works perferctly!
merlin
This should be the accepted answer. I came here twice after a few weeks since it is hard to remember and searched again for it.
Isn't this just a matter of splitting a string on a character?
>>> url = http://example.com/?a=text&q2=text2&q3=text3&q2=text4
>>> url = url.split('?')[0]
'http://example.com/'
4 Comments
Programer Beginner
I was thinking about this solution as well. Can anyone tell me if there are any issue (potential bug/loophole) in this proposed solution?
mevers303
@ProgramerBeginner There isn't one, really!
conradlee
The problem will be clear if you carefully read the original question. The OP wanted to remove only one parameter, all the query parameters.
Ahmed Djamel
This solution will work if all parameter was always sent in the same order, if you work with a URL that have unordered parameters you end deleting the wrong parameter.
Using python's url manipulation library furl:
import furl
f = furl.furl("http://example.com/?a=text&q2=text2&q3=text3&q2=text4")
f.remove(['q2'])
print(f.url)
1 Comment
Mattwmaster58
Calling it 'python's url manipulation library' makes it sound like it's included in the standard lib, which it isn't.
query_string = "https://example.com/api/api.php?user=chris&auth=true"
url = query_string[:query_string.find('?', 0)]
2 Comments
Alexander
this does not exactly provide a solution for the given answer. please try improving your answer or deleting it.
Nic3500
While this code may answer the question, providing additional context regarding how and/or why it solves the problem would improve the answer's long-term value.
Or simply put, just use url_query_cleaner() from w3lib.url
from w3lib.url import url_query_cleaner
url = 'http://example.com/?a=text&q2=text2&q3=text3&q2=text4'
url_query_cleaner(url, ('q2'), remove=True)
Output: http://example.com/?a=text&q3=text3
Comments
Another method that you can use to have more control over what you want to do is urlunparse() which takes a tuple of the parts returned from urlparse().
For example, recently I needed to change the path but keep the query:
from urllib.parse import urlparse, urlunparse
url = 'https://test.host.com/some/path?type_id=7'
parsed_url = urlparse(url)
modified_path = f'{parsed_url.path}/new_path_ending'
output_url = urlunparse((
parsed_url.scheme,
parsed_url.netloc,
modified_path,
parsed_url.params,
parsed_url.query,
parsed_url.fragment
))
print(output_url)
'https://test.host.com/some/path/new_path_ending?type_id=7'
This method preserves all of the URL and gives you granular control of what you want to keep, change, and remove.
Comments
import re
q ="http://example.com/?a=text&q2=text2&q3=text3&q2=text4"
todelete="q2"
#Delete every query string matching the pattern
r = re.sub(r''+todelete+'=[a-zA-Z_0-9]*\&*',r'',q)
#Delete the possible trailing #
r = re.sub(r'&$',r'',r)
print r
Comments
Or you could just use strip
>>> l='http://example.com/?a=text&q2=text2&q3=text3&q2=text4'
>>> l.strip('&q2=text4')
'http://example.com/?a=text&q2=text2&q3=text3'
>>>
