Retrieving parameters from a URL

Question

Given a URL like the following, how can I parse the value of the query parameters? For example, in this case I want the value of some_key .

/some_path?some_key=some_value'

I am using Django in my environment; is there a method on the request object that could help me?

I tried using self.request.get('some_key') but it is not returning the value some_value as I had hoped.

score 413 · Accepted Answer · edited Aug 26 '21 at 19:04

413

This is not specific to Django, but for Python in general. For a Django specific answer, see this one from @jball037

Python 2:

import urlparse

url = 'https://www.example.com/some_path?some_key=some_value'
parsed = urlparse.urlparse(url)
captured_value = urlparse.parse_qs(parsed.query)['some_key'][0]

print captured_value

Python 3:

from urllib.parse import urlparse
from urllib.parse import parse_qs

url = 'https://www.example.com/some_path?some_key=some_value'
parsed_url = urlparse(url)
captured_value = parse_qs(parsed_url.query)['some_key'][0]

print(captured_value)

parse_qs returns a list. The [0] gets the first item of the list so the output of each script is some_value

Here's the 'parse_qs' documentation for Python 3

edited Aug 26 '21 at 19:04

Alan W. Smith

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answered Feb 22 '11 at 07:54

systempuntoout

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In python3 `import urllib.parse as urlparse` – Ivan De Paz Centeno Sep 28 '16 at 09:44
34

Note that `parse_qs` will return you a list object. You need to get the first element of it if you want a string `urlparse.parse_qs(parsed.query)['def'][0]` – robertspierre Jul 09 '17 at 10:11
4

Note that if url contain '#' character just like this: http://foo.appspot.com/#/abc?def=ghi, you must disallow fragments: `urlparse(url, allow_fragments=False)`, otherwise query will return empty str. – codezjx Jul 13 '17 at 15:29
Py3 docs: [urlparse](https://docs.python.org/3/library/urllib.parse.html?highlight=urlparse#urllib.parse.urlparse); [parse_qs](https://docs.python.org/3/library/urllib.parse.html?highlight=parse_qs#urllib.parse.parse_qs). Py2 docs: [urlparse](https://docs.python.org/2.7/library/urlparse.html?highlight=urlparse#urlparse.urlparse); [parse_qs](https://docs.python.org/2.7/library/urlparse.html?highlight=parse_qs#urlparse.parse_qs). – Czechnology Aug 14 '18 at 14:09
5

in python3, must use `from urllib.parse import urlparse, parse_qs` and `parse_qs(parsed.query)` – deathemperor Aug 21 '19 at 11:37
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@FrancescoFrassinelli it's valid. `urlparse.urlparse(url)` --> `urlparse(url)` – Winand Jul 15 '20 at 07:22
@Winand you are right of course. I do not know why that evening I wrote that :) Thanks. – Francesco Frassinelli Jul 16 '20 at 08:24
What happens if the URL Is not valid (just a string) ... oR ... if the param is not present? – snh_nl Oct 30 '20 at 12:12
Getting AttributeError: 'ParseResult' object has no attribute 'decode', unsolved – Michael Mar 19 '21 at 02:02

score 91 · Answer 2 · edited Dec 18 '18 at 21:17

91

I'm shocked this solution isn't on here already. Use:

request.GET.get('variable_name')

This will "get" the variable from the "GET" dictionary, and return the 'variable_name' value if it exists, or a None object if it doesn't exist.

edited Dec 18 '18 at 21:17

Marcus Müller

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answered May 12 '14 at 17:30

jball037

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I can't get this to work. Maybe some more details. How are you getting request? Is this python3 compatible? – ggedde Apr 02 '19 at 20:51
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@ggedde this answer (like the question) is for django, it is a django request object. if you are not using django, use one of the other answers. – Florian Apr 14 '19 at 09:34
def parse_category_page(self, response): product_id = response.GET.get('number') doesnt work neither does request – snh_nl Sep 14 '20 at 07:36
I'm shocked that someone actually added this answer and has so many likes (just kidding ;) ). One thing is getting query params from a random URL, and a very different thing is getting those params from inside an endpoint that you are serving using Django or another framework. This answer is in the case your URL is precisely the one you are serving with Django. The question is about some random internet URL... should parse it instead. – juan Isaza Mar 28 '23 at 20:37

score 68 · Answer 3 · edited Feb 08 '18 at 01:40

68

for Python > 3.4

from urllib import parse
url = 'http://foo.appspot.com/abc?def=ghi'
query_def=parse.parse_qs(parse.urlparse(url).query)['def'][0]

edited Feb 08 '18 at 01:40

Rohan Sawant

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answered Jan 12 '17 at 10:43

Anatoly E

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Best way to do it IMO. Doesn't need extra packages and works A+ on Python 3.5. – wanaryytel Apr 06 '17 at 19:14

score 63 · Answer 4 · edited Jan 14 '19 at 02:51

63

import urlparse
url = 'http://example.com/?q=abc&p=123'
par = urlparse.parse_qs(urlparse.urlparse(url).query)

print par['q'][0], par['p'][0]

edited Jan 14 '19 at 02:51

Joonsoo

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answered Jan 31 '13 at 19:33

Cris

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score 35 · Answer 5 · answered Mar 17 '16 at 16:05

35

There is a new library called furl. I find this library to be most pythonic for doing url algebra. To install:

pip install furl

Code:

from furl import furl
f = furl("/abc?def='ghi'") 
print f.args['def']

answered Mar 17 '16 at 16:05

Mayank Jaiswal

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Great library! Didn't know about it but works like a charm :) – pasql Apr 07 '16 at 00:36
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:) I have wasted my share of time trying to make urlparse work for me. No more. – Mayank Jaiswal Apr 07 '16 at 07:04

score 21 · Answer 6 · answered Mar 01 '14 at 04:06

I know this is a bit late but since I found myself on here today, I thought that this might be a useful answer for others.

import urlparse
url = 'http://example.com/?q=abc&p=123'
parsed = urlparse.urlparse(url)
params = urlparse.parse_qsl(parsed.query)
for x,y in params:
    print "Parameter = "+x,"Value = "+y

With parse_qsl(), "Data are returned as a list of name, value pairs."

score 7 · Answer 7 · answered Aug 25 '17 at 16:28

7

In pure Python:

def get_param_from_url(url, param_name):
    return [i.split("=")[-1] for i in url.split("?", 1)[-1].split("&") if i.startswith(param_name + "=")][0]

answered Aug 25 '17 at 16:28

Asym

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score 7 · Answer 8 · answered Feb 22 '11 at 06:40

7

The url you are referring is a query type and I see that the request object supports a method called arguments to get the query arguments. You may also want try self.request.get('def') directly to get your value from the object..

answered Feb 22 '11 at 06:40

Senthil Kumaran

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score 6 · Answer 9 · answered Feb 22 '11 at 06:44

6

def getParams(url):
    params = url.split("?")[1]
    params = params.split('=')
    pairs = zip(params[0::2], params[1::2])
    answer = dict((k,v) for k,v in pairs)

Hope this helps

answered Feb 22 '11 at 06:44

inspectorG4dget

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This should not be necessary inside a webapp. – Nick Johnson Feb 23 '11 at 03:48

score 4 · Answer 10 · answered Oct 05 '14 at 01:14

There's not need to do any of that. Only with

self.request.get('variable_name')

Notice that I'm not specifying the method (GET, POST, etc). This is well documented and this is an example

The fact that you use Django templates doesn't mean the handler is processed by Django as well

score 3 · Answer 11 · answered Feb 22 '11 at 06:59

3

The urlparse module provides everything you need:

urlparse.parse_qs()

answered Feb 22 '11 at 06:59

Does this answer provide anything not provided by other answers? – Joshua Snider Aug 11 '21 at 09:10

score 3 · Answer 12 · answered Aug 16 '20 at 22:36

Most answers here suggest using parse_qs to parse an URL string. This method always returns the values as a list (not directly as a string) because a parameter can appear multiple times, e.g.:

http://example.com/?foo=bar&foo=baz&bar=baz

Would return:

{'foo': ['bar', 'baz'], 'bar' : ['baz']}

This is a bit inconvenient because in most cases you're dealing with an URL that doesn't have the same parameter multiple times. This function returns the first value by default, and only returns a list if there's more than one element.

from urllib import parse

def parse_urlargs(url):
    query = parse.parse_qs(parse.urlparse(url).query)
    return {k:v[0] if v and len(v) == 1 else v for k,v in query.items()}

For example, http://example.com/?foo=bar&foo=baz&bar=baz would return:

{'foo': ['bar', 'baz'], 'bar': 'baz'}

score 3 · Answer 13 · edited Mar 08 '14 at 21:56

3

import cgitb
cgitb.enable()

import cgi
print "Content-Type: text/plain;charset=utf-8"
print
form = cgi.FieldStorage()
i = int(form.getvalue('a'))+int(form.getvalue('b'))
print i

edited Mar 08 '14 at 21:56

Nate C-K

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answered Jan 21 '12 at 16:01

George

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score 2 · Answer 14 · answered Sep 12 '17 at 15:25

2

There is a nice library w3lib.url

from w3lib.url import url_query_parameter
url = "/abc?def=ghi"
print url_query_parameter(url, 'def')
ghi

answered Sep 12 '17 at 15:25

Corvax

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score 2 · Answer 15 · answered Aug 21 '22 at 23:11

2

Parsing it in raw python:

path = '/some_path?k1=v1&k2=v2&k3=v3'
_, query_string = path.split('?')
params  = dict(param.split('=') for param in query_string.split('&'))
print(params)

Output:

{'k1': 'v1', 'k2': 'v2', 'k3': 'v3'}

answered Aug 21 '22 at 23:11

haccks

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score 0 · Answer 16 · answered Jul 25 '12 at 00:31

Btw, I was having issues using parse_qs() and getting empty value parameters and learned that you have to pass a second optional parameter 'keep_blank_values' to return a list of the parameters in a query string that contain no values. It defaults to false. Some crappy written APIs require parameters to be present even if they contain no values

for k,v in urlparse.parse_qs(p.query, True).items():
  print k

score 0 · Answer 17 · answered Nov 11 '19 at 17:10

0

parameters = dict([part.split('=') for part in get_parsed_url[4].split('&')])

This one is simple. The variable parameters will contain a dictionary of all the parameters.

answered Nov 11 '19 at 17:10

Vivek

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score 0 · Answer 18 · answered Jan 14 '20 at 15:41

I see there isn't an answer for users of Tornado:

key = self.request.query_arguments.get("key", None)

This method must work inside an handler that is derived from:

tornado.web.RequestHandler

None is the answer this method will return when the requested key can't be found. This saves you some exception handling.

curveball · Answer 19 · 2020-09-16T23:57:58.163

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I didn't want to mess with additional libraries. Simple ways suggested here didn't work out either. Finally, not on the request object, but I could get a GET parameter w/o all that hassle via self.GET.get('XXX'):

...
def get_context_data(self, **kwargs):
    context = super(SomeView, self).get_context_data(**kwargs)
    context['XXX'] = self.GET.get('XXX')
...

Python 2.7.18, Django 1.11.20

edited Sep 16 '20 at 23:57

answered Sep 16 '20 at 20:41

curveball

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score 0 · Answer 20 · answered Dec 22 '21 at 10:57

NOT using Django and tried on Python 3.9. It can also be retrieved using the below code snippet in the handler:

considering URL as-

http://localhost:8081/api/v1/users?query_key=abcdef12345

and handler method as:

@routes.get('/api/v1/users')
async def handler_users(request):
    query_key = request.url.query['query_key']

score 0 · Answer 21 · answered Mar 04 '23 at 16:49

0

You can use only request.GET.get() method

urls.py:

    path('some_path', views.func),

views.py:

def func(request):
    some_key = request.GET.get('some_key', None)
    ...

answered Mar 04 '23 at 16:49

phaestos

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Retrieving parameters from a URL

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