python - 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 def.

/abc?def='ghi'

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('def') but it is not returning the value ghi as I had hoped.

Answer 1

Python 2:

import urlparse
url = 'http://foo.appspot.com/abc?def=ghi'
parsed = urlparse.urlparse(url)
print urlparse.parse_qs(parsed.query)['def']

Python 3:

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

parse_qs returns a list of values, so the above code will print ['ghi'].

Here's the Python 3 documentation.

Answer 2

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.

Answer 3

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]

Answer 4

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]

Answer 5

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']

Answer 6

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."

Answer 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..

Answer 8

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

Answer 9

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]

Answer 10

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

Answer 11

The urlparse module provides everything you need:

urlparse.parse_qs()

Answer 12

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

Answer 13

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'}

Answer 14

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

Answer 15

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

Answer 16

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.

Answer 17

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.

Answer 18

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