Add params to given URL in Python
Suppose I was given a URL.
It might already have GET parameters (e.g.http://example.com/search?q=question
) or it might not (e.g. http://example.com/
).
And now I need to add some parameters to it like {'lang':'en','tag':'python'}
. In the first case I'm going to have http://example.com/search?q=question&lang=en&开发者_如何学编程tag=python
and in the second — http://example.com/search?lang=en&tag=python
.
Is there any standard way to do this?
There are a couple of quirks with the urllib
and urlparse
modules. Here's a working example:
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
, the result of urlparse()
, is read-only and we need to convert it to a list
before we can attempt to modify its data.
Outsource it to the battle tested requests library.
This is how I will do it:
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)
Why
I've been not satisfied with all the solutions on this page (come on, where is our favorite copy-paste thing?) so I wrote my own based on answers here. It tries to be complete and more Pythonic. I've added a handler for dict and bool values in arguments to be more consumer-side (JS) friendly, but they are yet optional, you can drop them.
How it works
Test 1: Adding new arguments, handling Arrays and Bool values:
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'
Test 2: Rewriting existing args, handling DICT values:
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'
Talk is cheap. Show me the code.
Code itself. I've tried to describe it in details:
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
Please be aware that there may be some issues, if you'll find one please let me know and we will make this thing better
You want to use URL encoding if the strings can have arbitrary data (for example, characters such as ampersands, slashes, etc. will need to be encoded).
Check out urllib.urlencode:
>>> import urllib
>>> urllib.urlencode({'lang':'en','tag':'python'})
'lang=en&tag=python'
In python3:
from urllib import parse
parse.urlencode({'lang':'en','tag':'python'})
You can also use the furl module 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
If you are using the requests lib:
import requests
...
params = {'tag': 'python'}
requests.get(url, params=params)
Based on this answer, one-liner for simple cases (Python 3 code):
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)
or:
url += ('&', '?')[urlparse(url).query == ''] + urlencode(params)
I find this more elegant than the two top answers:
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'
The most important things I dislike in the top answers (they are nevertheless good):
- Łukasz: having to remember the index at which the
query
is in the URL components - Sapphire64: the very verbose way of creating the updated
ParseResult
What's bad about my response is the magically looking dict
merge using unpacking, but I prefer that to updating an already existing dictionary because of my prejudice against mutability.
Yes: use urllib.
From the examples in the documentation:
>>> 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
python3
, self explanatory I guess
from urllib.parse import urlparse, urlencode, parse_qsl
url = 'https://www.linkedin.com/jobs/search?keywords=engineer'
parsed = urlparse(url)
current_params = dict(parse_qsl(parsed.query))
new_params = {'location': 'United States'}
merged_params = urlencode({**current_params, **new_params})
parsed = parsed._replace(query=merged_params)
print(parsed.geturl())
# https://www.linkedin.com/jobs/search?keywords=engineer&location=United+States
I liked Łukasz version, but since urllib and urllparse functions are somewhat awkward to use in this case, I think it's more straightforward to do something like this:
params = urllib.urlencode(params)
if urlparse.urlparse(url)[4]:
print url + '&' + params
else:
print url + '?' + params
Use the various urlparse
functions to tear apart the existing URL, urllib.urlencode()
on the combined dictionary, then urlparse.urlunparse()
to put it all back together again.
Or just take the result of urllib.urlencode()
and concatenate it to the URL appropriately.
Yet another answer:
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))
In python 2.5
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"
Here is how I implemented it.
import urllib
params = urllib.urlencode({'lang':'en','tag':'python'})
url = ''
if request.GET:
url = request.url + '&' + params
else:
url = request.url + '?' + params
Worked like a charm. However, I would have liked a more cleaner way to implement this.
Another way of implementing the above is put it in a method.
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
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