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字典python的URL查询参数

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问题：字典python的URL查询参数
回答 0
回答 1
回答 2
回答 3
回答 4
回答 5

问题：字典python的URL查询参数

有没有一种方法可以解析网址（带有某些python库）并返回带有该网址查询参数部分的键和值的python字典？

例如：

url = "http://www.example.org/default.html?ct=32&op=92&item=98"

预期收益：

{'ct':32, 'op':92, 'item':98}
点击查看英文原文

Is there a way to parse a URL (with some python library) and return a python dictionary with the keys and values of a query parameters part of the URL?

For example:

url = "http://www.example.org/default.html?ct=32&op=92&item=98"

expected return:

{'ct':32, 'op':92, 'item':98}

回答 0

使用urllib.parse库：

>>> from urllib import parse
>>> url = "http://www.example.org/default.html?ct=32&op=92&item=98"
>>> parse.urlsplit(url)
SplitResult(scheme='http', netloc='www.example.org', path='/default.html', query='ct=32&op=92&item=98', fragment='')
>>> parse.parse_qs(parse.urlsplit(url).query)
{'item': ['98'], 'op': ['92'], 'ct': ['32']}
>>> dict(parse.parse_qsl(parse.urlsplit(url).query))
{'item': '98', 'op': '92', 'ct': '32'}

在urllib.parse.parse_qs()与urllib.parse.parse_qsl()方法解析出查询字符串，考虑到钥匙可能会出现不止一次和顺序可能无关紧要。

如果您仍在使用Python 2，urllib.parse则称为urlparse。

点击查看英文原文

Use the urllib.parse library:

>>> from urllib import parse
>>> url = "http://www.example.org/default.html?ct=32&op=92&item=98"
>>> parse.urlsplit(url)
SplitResult(scheme='http', netloc='www.example.org', path='/default.html', query='ct=32&op=92&item=98', fragment='')
>>> parse.parse_qs(parse.urlsplit(url).query)
{'item': ['98'], 'op': ['92'], 'ct': ['32']}
>>> dict(parse.parse_qsl(parse.urlsplit(url).query))
{'item': '98', 'op': '92', 'ct': '32'}

The urllib.parse.parse_qs() and urllib.parse.parse_qsl() methods parse out query strings, taking into account that keys can occur more than once and that order may matter.

If you are still on Python 2, urllib.parse was called urlparse.

回答 1

对于Python 3，dict from的值parse_qs在列表中，因为可能有多个值。如果您只想要第一个：

>>> from urllib.parse import urlsplit, parse_qs
>>>
>>> url = "http://www.example.org/default.html?ct=32&op=92&item=98"
>>> query = urlsplit(url).query
>>> params = parse_qs(query)
>>> params
{'item': ['98'], 'op': ['92'], 'ct': ['32']}
>>> dict(params)
{'item': ['98'], 'op': ['92'], 'ct': ['32']}
>>> {k: v[0] for k, v in params.items()}
{'item': '98', 'op': '92', 'ct': '32'}
点击查看英文原文

For Python 3, the values of the dict from parse_qs are in a list, because there might be multiple values. If you just want the first one:

>>> from urllib.parse import urlsplit, parse_qs
>>>
>>> url = "http://www.example.org/default.html?ct=32&op=92&item=98"
>>> query = urlsplit(url).query
>>> params = parse_qs(query)
>>> params
{'item': ['98'], 'op': ['92'], 'ct': ['32']}
>>> dict(params)
{'item': ['98'], 'op': ['92'], 'ct': ['32']}
>>> {k: v[0] for k, v in params.items()}
{'item': '98', 'op': '92', 'ct': '32'}

回答 2

如果您不想使用解析器：

url = "http://www.example.org/default.html?ct=32&op=92&item=98"
url = url.split("?")[1]
dict = {x[0] : x[1] for x in [x.split("=") for x in url[1:].split("&") ]}

因此，我不会删除上面的内容，但是绝对不是您应该使用的内容。

我想我读了一些答案，而且它们看起来有些复杂，以防万一您像我一样，不要使用我的解决方案。

用这个：

from urllib import parse
params = dict(parse.parse_qsl(parse.urlsplit(url).query))

而对于Python 2.X

import urlparse as parse
params = dict(parse.parse_qsl(parse.urlsplit(url).query))

我知道这与接受的答案相同，只是在一个可以复制的衬里上。

点击查看英文原文

If you prefer not to use a parser:

url = "http://www.example.org/default.html?ct=32&op=92&item=98"
url = url.split("?")[1]
dict = {x[0] : x[1] for x in [x.split("=") for x in url[1:].split("&") ]}

So I won’t delete what’s above but it’s definitely not what you should use.

I think I read a few of the answers and they looked a little complicated, incase you’re like me, don’t use my solution.

Use this:

from urllib import parse
params = dict(parse.parse_qsl(parse.urlsplit(url).query))

and for Python 2.X

import urlparse as parse
params = dict(parse.parse_qsl(parse.urlsplit(url).query))

I know this is the same as the accepted answer, just in a one liner that can be copied.

回答 3

对于python 2.7

In [14]: url = "http://www.example.org/default.html?ct=32&op=92&item=98"

In [15]: from urlparse import urlparse, parse_qsl

In [16]: parse_url = urlparse(url)

In [17]: query_dict = dict(parse_qsl(parse_url.query))

In [18]: query_dict
Out[18]: {'ct': '32', 'item': '98', 'op': '92'}
点击查看英文原文

For python 2.7

In [14]: url = "http://www.example.org/default.html?ct=32&op=92&item=98"

In [15]: from urlparse import urlparse, parse_qsl

In [16]: parse_url = urlparse(url)

In [17]: query_dict = dict(parse_qsl(parse_url.query))

In [18]: query_dict
Out[18]: {'ct': '32', 'item': '98', 'op': '92'}

回答 4

我同意不重新发明轮子，但有时（在您学习时）有助于构建轮子以便理解轮子。:)因此，从纯粹的学术角度来看，我提供了一个警告，即使用字典假定名称/值对是唯一的（查询字符串不包含多个记录）。

url = 'http:/mypage.html?one=1&two=2&three=3'

page, query = url.split('?')

names_values_dict = dict(pair.split('=') for pair in query.split('&'))

names_values_list = [pair.split('=') for pair in query.split('&')]

我在空闲IDE中使用3.6.5版。

点击查看英文原文

I agree about not reinventing the wheel but sometimes (while you’re learning) it helps to build a wheel in order to understand a wheel. :) So, from a purely academic perspective, I offer this with the caveat that using a dictionary assumes that name value pairs are unique (that the query string does not contain multiple records).

url = 'http:/mypage.html?one=1&two=2&three=3'

page, query = url.split('?')

names_values_dict = dict(pair.split('=') for pair in query.split('&'))

names_values_list = [pair.split('=') for pair in query.split('&')]

I’m using version 3.6.5 in the Idle IDE.

回答 5

对于python2.7我正在使用urlparse模块来解析URL查询到字典。

import urlparse

url = "http://www.example.org/default.html?ct=32&op=92&item=98"

print urlparse.parse_qs( urlparse.urlparse(url).query )
# result: {'item': ['98'], 'op': ['92'], 'ct': ['32']}
点击查看英文原文

For python2.7 I am using urlparse module to parse url query to dict. 

import urlparse

url = "http://www.example.org/default.html?ct=32&op=92&item=98"

print urlparse.parse_qs( urlparse.urlparse(url).query )
# result: {'item': ['98'], 'op': ['92'], 'ct': ['32']}