在Python中将params添加到给定的URL

假设我获得了一个URL。
它可能已经有GET参数(例如{'lang':'en','tag':'python'})或者它可能没有(例如http://example.com/search?q=question&lang=en&tag=python)。

现在我需要添加一些参数,如{'lang':'en','tag':'python'}.在第一种情况下,我将有http://example.com/search?q=question&lang=en&tag=python,在第二种情况下 - http://example.com/search?lang=en&tag=python

有没有标准的方法来做到这一点?

z4y4ts asked 2019-08-13T03:24:58Z
14个解决方案
152 votes

ParseResulturlparse()模块存在一些怪癖。 这是一个有效的例子:

try:
    import urlparse
    from urllib import urlencode
except: # For Python 3
    import urllib.parse as urlparse
    from urllib.parse import urlencode

url = "http://stackoverflow.com/search?q=question"
params = {'lang':'en','tag':'python'}

url_parts = list(urlparse.urlparse(url))
query = dict(urlparse.parse_qsl(url_parts[4]))
query.update(params)

url_parts[4] = urlencode(query)

print(urlparse.urlunparse(url_parts))

ParseResult,2484848727230622977的结果是只读的,我们需要将其转换为list才能尝试修改其数据。

Łukasz answered 2019-08-13T03:25:19Z
38 votes

为什么

我对这个页面上的所有解决方案都不满意(来吧,我们最喜欢的复制粘贴的东西在哪里?)所以我根据这里的答案编写了自己的解决方案。 它试图完成并且更加Pythonic。 我在参数中为dict和bool值添加了一个处理程序,以便更加消费者(JS)友好,但它们是可选的,你可以放弃它们。

这个怎么运作

测试1:添加新参数,处理Arrays和Bool值:

url = 'http://stackoverflow.com/test'
new_params = {'answers': False, 'data': ['some','values']}

add_url_params(url, new_params) == \
    'http://stackoverflow.com/test?data=some&data=values&answers=false'

测试2:重写现有的args,处理DICT值:

url = 'http://stackoverflow.com/test/?question=false'
new_params = {'question': {'__X__':'__Y__'}}

add_url_params(url, new_params) == \
    'http://stackoverflow.com/test/?question=%7B%22__X__%22%3A+%22__Y__%22%7D'

谈话很便宜。 告诉我代码。

代码本身。 我试图详细描述它:

from json import dumps

try:
    from urllib import urlencode, unquote
    from urlparse import urlparse, parse_qsl, ParseResult
except ImportError:
    # Python 3 fallback
    from urllib.parse import (
        urlencode, unquote, urlparse, parse_qsl, ParseResult
    )


def add_url_params(url, params):
    """ Add GET params to provided URL being aware of existing.

    :param url: string of target URL
    :param params: dict containing requested params to be added
    :return: string with updated URL

    >> url = 'http://stackoverflow.com/test?answers=true'
    >> new_params = {'answers': False, 'data': ['some','values']}
    >> add_url_params(url, new_params)
    'http://stackoverflow.com/test?data=some&data=values&answers=false'
    """
    # Unquoting URL first so we don't loose existing args
    url = unquote(url)
    # Extracting url info
    parsed_url = urlparse(url)
    # Extracting URL arguments from parsed URL
    get_args = parsed_url.query
    # Converting URL arguments to dict
    parsed_get_args = dict(parse_qsl(get_args))
    # Merging URL arguments dict with new params
    parsed_get_args.update(params)

    # Bool and Dict values should be converted to json-friendly values
    # you may throw this part away if you don't like it :)
    parsed_get_args.update(
        {k: dumps(v) for k, v in parsed_get_args.items()
         if isinstance(v, (bool, dict))}
    )

    # Converting URL argument to proper query string
    encoded_get_args = urlencode(parsed_get_args, doseq=True)
    # Creating new parsed result object based on provided with new
    # URL arguments. Same thing happens inside of urlparse.
    new_url = ParseResult(
        parsed_url.scheme, parsed_url.netloc, parsed_url.path,
        parsed_url.params, encoded_get_args, parsed_url.fragment
    ).geturl()

    return new_url

请注意,可能存在一些问题,如果您发现问题,请告诉我,我们会更好地做这件事

Sapphire64 answered 2019-08-13T03:26:24Z
31 votes

如果字符串可以包含任意数据,则需要使用URL编码(例如,“&符”,“斜杠”等字符需要进行编码)。

查看urllib.urlencode:

>>> import urllib
>>> urllib.urlencode({'lang':'en','tag':'python'})
'lang=en&tag=python'
Mike Mueller answered 2019-08-13T03:26:56Z
17 votes

你也可以使用furl模块[https://github.com/gruns/furl]

>>> from furl import furl
>>> print furl('http://example.com/search?q=question').add({'lang':'en','tag':'python'}).url
http://example.com/search?q=question&lang=en&tag=python
surfeurX answered 2019-08-13T03:27:22Z
9 votes

是的:使用urllib。

从文档中的示例:

>>> import urllib
>>> params = urllib.urlencode({'spam': 1, 'eggs': 2, 'bacon': 0})
>>> f = urllib.urlopen("http://www.musi-cal.com/cgi-bin/query?%s" % params)
>>> print f.geturl() # Prints the final URL with parameters.
>>> print f.read() # Prints the contents
unwind answered 2019-08-13T03:27:55Z
7 votes

根据这个答案,简单案例的单线程(Python 3代码):

from urllib.parse import urlparse, urlencode


url = "https://stackoverflow.com/search?q=question"
params = {'lang':'en','tag':'python'}

url += ('&' if urlparse(url).query else '?') + urlencode(params)

要么:

url += ('&', '?')[urlparse(url).query == ''] + urlencode(params)
Mikhail Gerasimov answered 2019-08-13T03:28:22Z
7 votes

如果您正在使用请求lib:

import requests
...
params = {'tag': 'python'}
requests.get(url, params=params)
Christophe Roussy answered 2019-08-13T03:28:47Z
6 votes

我喜欢Łukasz版本,但由于urllib和urllparse函数在这种情况下使用起来有些尴尬,我认为做这样的事情更简单:

params = urllib.urlencode(params)

if urlparse.urlparse(url)[4]:
    print url + '&' + params
else:
    print url + '?' + params
Facundo Olano answered 2019-08-13T03:29:12Z
5 votes

将其外包给经过测试的战斗请求库。

我就是这样做的:

from requests.models import PreparedRequest
url = 'http://example.com/search?q=question'
params = {'lang':'en','tag':'python'}
req = PreparedRequest()
req.prepare_url(url, params)
print(req.url)
Varun answered 2019-08-13T03:29:45Z
3 votes

使用各种urllib.urlencode()函数拆分现有的URL,组合字典上的urllib.urlencode(),然后再将urllib.urlencode()重新组合在一起。

或者只是获取urllib.urlencode()的结果并将其连接到URL。

Ignacio Vazquez-Abrams answered 2019-08-13T03:30:18Z
3 votes

还有一个答案:

def addGetParameters(url, newParams):
    (scheme, netloc, path, params, query, fragment) = urlparse.urlparse(url)
    queryList = urlparse.parse_qsl(query, keep_blank_values=True)
    for key in newParams:
        queryList.append((key, newParams[key]))
    return urlparse.urlunparse((scheme, netloc, path, params, urllib.urlencode(queryList), fragment))
Timmmm answered 2019-08-13T03:30:39Z
2 votes

我觉得这比两个顶级答案更优雅:

from urllib.parse import urlencode, urlparse, parse_qs

def merge_url_query_params(url: str, additional_params: dict) -> str:
    url_components = urlparse(url)
    original_params = parse_qs(url_components.query)
    # Before Python 3.5 you could update original_params with 
    # additional_params, but here all the variables are immutable.
    merged_params = {**original_params, **additional_params}
    updated_query = urlencode(merged_params, doseq=True)
    # _replace() is how you can create a new NamedTuple with a changed field
    return url_components._replace(query=updated_query).geturl()

assert merge_url_query_params(
    'http://example.com/search?q=question',
    {'lang':'en','tag':'python'},
) == 'http://example.com/search?q=question&lang=en&tag=python'

在最顶层的答案中我最不重要的事情(它们仍然很好):

  • Łukasz:必须记住query在URL组件中的索引
  • Sapphire64:创建更新dict的非常冗长的方式

我的回答有什么不好的地方是使用解压缩看起来神奇的dict合并,但由于我对可变性的偏见,我更喜欢更新已经存在的字典。

butla answered 2019-08-13T03:31:34Z
1 votes

在python 2.5中

import cgi
import urllib
import urlparse

def add_url_param(url, **params):
    n=3
    parts = list(urlparse.urlsplit(url))
    d = dict(cgi.parse_qsl(parts[n])) # use cgi.parse_qs for list values
    d.update(params)
    parts[n]=urllib.urlencode(d)
    return urlparse.urlunsplit(parts)

url = "http://stackoverflow.com/search?q=question"
add_url_param(url, lang='en') == "http://stackoverflow.com/search?q=question&lang=en"
dpatru answered 2019-08-13T03:32:01Z
1 votes

这是我实现它的方式。

import urllib

params = urllib.urlencode({'lang':'en','tag':'python'})
url = ''
if request.GET:
   url = request.url + '&' + params
else:
   url = request.url + '?' + params    

工作就像一个魅力。 但是,我希望有一种更清洁的方式来实现它。

实现上述内容的另一种方法是将其置于一种方法中。

import urllib

def add_url_param(request, **params):
   new_url = ''
   _params = dict(**params)
   _params = urllib.urlencode(_params)

   if _params:
      if request.GET:
         new_url = request.url + '&' + _params
      else:
         new_url = request.url + '?' + _params
   else:
      new_url = request.url

   return new_ur
Monty answered 2019-08-13T03:32:41Z
translate from https://stackoverflow.com:/questions/2506379/add-params-to-given-url-in-python