如何使用Python从URL中删除查询字符串

Question

示例：

http://example.com/?a=text&q2=text2&q3=text3&q2=text4

删除“ q2”后，将返回：

http://example.com/?q=text&q3=text3

在这种情况下，有多个“ q2”，并且都已删除。

Answer 1

import cgi
import urlparse
url = "http://example.com/?a=text&q2=text2&q3=text3&q2=text4"
qs = cgi.parse_qs(urlparse.urlparse(url)[4])
del(qs['q2'])
print qs

明确地...

>>> import cgi
>>> import urlparse
>>> url = "http://example.com/?a=text&q2=text2&q3=text3&q2=text4"
>>> qs = cgi.parse_qs(urlparse.urlparse(url)[4])
>>> del(qs['q2'])
>>> print qs
{'a': ['text'], 'q3': ['text3']}
>>>

Answer 2

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))

输出：

http://example.com/?a=text&q3=text3&b=#q2=keep_fragment

Answer 3

import urlparse

url = 'http://example.com/?a=text&q2=text2&q3=text3&q2=text4'
urlparse.urljoin(url, urlparse.urlparse(url).path)

Answer 4

使用python的url操作库furl：

import furl
f = furl.furl("http://example.com/?a=text&q2=text2&q3=text3&q2=text4")
f.remove(['q2'])
print(f.url)

Answer 5

这不只是在字符上分割字符串的问题吗？

>>> url = http://example.com/?a=text&q2=text2&q3=text3&q2=text4
>>> url = url.split('?')[0]
'http://example.com/'

Answer 6

query_string = "https://example.com/api/api.php?user=chris&auth=true"
url = query_string[:query_string.find('?', 0)]

Answer 7

或者简单地说，只需使用url_query_cleaner()中的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)

输出：http://example.com/?a=text&q3=text3

Answer 8

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