Python Dictionary to URL Parameters

Question

I am trying to convert a Python dictionary to a string for use as URL parameters. I am sure that there is a better, more Pythonic way of doing this. What is it?

x = ""
for key, val in {'a':'A', 'b':'B'}.items():
    x += "%s=%s&" %(key,val)
x = x[:-1]

Innis

Answer 1

Here is the correct way of using it in Python 3.

from urllib.parse import urlencode
params = {'a':'A', 'b':'B'}
print(urlencode(params))

Wallford

Since python 2 is now deprecated I would accept this answer but stack overflow is preventing me from changing the accepted answer. – Innis

How is this different from my answer which was ~ a year earlier before this one? – Slipsheet

Answer 2

Use urllib.parse.urlencode(). It takes a dictionary of key-value pairs, and converts it into a form suitable for a URL (e.g., key1=val1&key2=val2).

For your example:

>>> import urllib.parse
>>> params = {'a':'A', 'b':'B'}
>>> urllib.parse.urlencode(params)
'a=A&b=B'

If you want to make a URL with repetitive params such as: p=1&p=2&p=3 you have two options:

>>> a = (('p',1),('p',2), ('p', 3))
>>> urllib.parse.urlencode(a)
'p=1&p=2&p=3'

or:

>>> urllib.parse.urlencode({'p': [1, 2, 3]}, doseq=True)
'p=1&p=2&p=3'

If you are still using Python 2, use urllib.urlencode().

Hubing

if you want to make a url with repetitive params for example: ?p=1&p=2&p=3 then a = (('p',1),('p',2), ('p', 3)); urllib.urlencode(a) the result is 'p=1&p=2&p=3' – Febri

Another way to get repetitive params: urllib.urlencode({'p': [1, 2, 3]}, doseq=True) resulting in 'p=1&p=2&p=3'. – Swearword

If you wonder what doeseq is about: "If any values in the query arg are sequences and doseq is true, each sequence element is converted to a separate parameter." – Chivaree

Python3 users: urllib.parse.urlencode() – Witha

Answer 3

For python 3, the urllib library has changed a bit, now you have to do:

from urllib.parse import urlencode


params = {'a':'A', 'b':'B'}

urlencode(params)

Slipsheet

Answer 4

Use the 3rd party Python url manipulation library furl:

f = furl.furl('')
f.args = {'a':'A', 'b':'B'}
print(f.url) # prints ... '?a=A&b=B'

If you want repetitive parameters, you can do the following:

f = furl.furl('')
f.args = [('a', 'A'), ('b', 'B'),('b', 'B2')]
print(f.url) # prints ... '?a=A&b=B&b=B2'

Certes

Where do I get furl? It appears not to be a standard library – Riant

pip install furl Its not a part of standard library – Certes

Answer 5

This seems a bit more Pythonic to me, and doesn't use any other modules:

x = '&'.join(["{}={}".format(k, v) for k, v in {'a':'A', 'b':'B'}.items()])

Arv

This won't percent encode the parameters properly. This will create unexpected results if your data includes ampersands, equals, hash symbols, etc. – Sawmill