162

After using cgi.parse_qs(), how to convert the result (dictionary) back to query string? Looking for something similar to urllib.urlencode().

dreftymac
  • 31,404
  • 26
  • 119
  • 182
Sam
  • 4,727
  • 9
  • 23
  • 17
  • 7
    N.B: [`cgi.parse_qs()` is deprecated. Use urlparse.parse_qs() instead.](http://docs.python.org/library/cgi.html#cgi.parse_qs) – johnsyweb Oct 18 '11 at 04:53
  • 1
    **See also:** https://stackoverflow.com/questions/5607551/how-to-urlencode-a-querystring-in-python – dreftymac Jul 13 '17 at 19:59

4 Answers4

214

Python 3

urllib.parse.urlencode(query, doseq=False, [...])

Convert a mapping object or a sequence of two-element tuples, which may contain str or bytes objects, to a percent-encoded ASCII text string.

— Python 3 urllib.parse docs

A dict is a mapping.

Legacy Python

urllib.urlencode(query[, doseq])
Convert a mapping object or a sequence of two-element tuples to a “percent-encoded” string... a series of key=value pairs separated by '&' characters...

— Python 2.7 urllib docs

aschmid00
  • 7,038
  • 2
  • 47
  • 66
Ignacio Vazquez-Abrams
  • 776,304
  • 153
  • 1,341
  • 1,358
  • 1
    This is true, but the `dict` returned by `cgi.parse_qs()` actually has `list`s as its "values". Passing these straight in will result in very odd looking query strings. – johnsyweb Oct 18 '11 at 04:50
  • True enough. That'll teach me to read more than the first sentence! – johnsyweb Oct 18 '11 at 04:55
  • 18
    [urllib.parse.urlencode](https://docs.python.org/3.3/library/urllib.parse.html#urllib.parse.urlencode) in Python 3. – Denilson Sá Maia Oct 17 '14 at 21:41
  • 1
    The problem is that urlencode will convert space into +, which is not recommended. – user1633272 May 17 '17 at 06:46
  • 3
    @user1633272: Not recommended according to who? – Ignacio Vazquez-Abrams May 17 '17 at 07:46
  • is there a way that I can use `urlencode()` just to make use of `doseq` param. i.e. I do not want any of the characters to be encoded. I know there is a `safe` param, but that would require me to list out each character. I am looking for a wild-card character for `safe` param so that it understands that it has to encode nothing. – Vikas Prasad Sep 09 '20 at 06:35
  • Based on my findings, I do not think anything like such^ exists. And thus, I am exploring the option to first let it get encode and then decode. I know its a round-trip, but I couldn't think of any other alternative except writing the logic of `doseq` on my own in my codebase. – Vikas Prasad Sep 09 '20 at 06:40
  • I realised I can achieve this^^ by passing my own `quote_via` function. Something like: `urlencode(params, doseq=True, quote_via=lambda a, b, c, d : a)` – Vikas Prasad Sep 10 '20 at 04:33
  • Instead of `unquote_plus(urlencode(params, doseq=True))` – Vikas Prasad Sep 10 '20 at 06:25
107

In python3, slightly different:

from urllib.parse import urlencode
urlencode({'pram1': 'foo', 'param2': 'bar'})

output: 'pram1=foo&param2=bar'

for python2 and python3 compatibility, try this:

try:
    #python2
    from urllib import urlencode
except ImportError:
    #python3
    from urllib.parse import urlencode
BioGeek
  • 21,897
  • 23
  • 83
  • 145
Johnny Zhao
  • 2,858
  • 2
  • 29
  • 26
12

You're looking for something exactly like urllib.urlencode()!

However, when you call parse_qs() (distinct from parse_qsl()), the dictionary keys are the unique query variable names and the values are lists of values for each name.

In order to pass this information into urllib.urlencode(), you must "flatten" these lists. Here is how you can do it with a list comprehenshion of tuples: 

query_pairs = [(k,v) for k,vlist in d.iteritems() for v in vlist]
urllib.urlencode(query_pairs)
johnsyweb
  • 136,902
  • 23
  • 188
  • 247
1

Maybe you're looking for something like this:

def dictToQuery(d):
  query = ''
  for key in d.keys():
    query += str(key) + '=' + str(d[key]) + "&"
  return query

It takes a dictionary and convert it to a query string, just like urlencode. It'll append a final "&" to the query string, but return query[:-1] fixes that, if it's an issue.

garbados
  • 885
  • 5
  • 21
  • 4
    Have you met `str.join()` yet? How about `urllib.quote_plus()`? – Ignacio Vazquez-Abrams Oct 18 '11 at 04:34
  • 1
    @garbados Your solution should work in simple cases. However, have a look `urlencode` in `urllib.py` (it should by in your Python install) to see why creating a query string is sometimes not quite as simple as your answer implies (in particular the need to 'quote' certain characters that aren't valid in a URL). @Ignacio has also referenced two functions that would clean up your implementation and make it correct. – Anthony Cramp Oct 18 '11 at 04:57
  • Ah, indeed. Sorry for the cruddy implementation. Goes to show I should be more careful answering questions I've never faced myself. – garbados Oct 20 '11 at 06:29
  • While this may not be the best answer, it is exactly what I wanted to see when I clicked on this question's title. I have only 4 items in a dict that I need to turn into a name=value pair separated by an '&'. My values are controlled. The str.join() is handy in my case but I have no need for quote_plus(), again, because it's not public-facing code :) Thanks! – harperville Apr 09 '14 at 16:55
  • 1
    @harperville but the more correct way (`from urllib.parse import urlencode; urlencode(your_dict)`) is shorter and easier than this! I'll grant that it's sometimes smart to reinvent the wheel, even shoddily, when it's expensive or inconvenient to access existing, well-designed wheels, but here using the off-the-shelf wheel is easier and quicker than rolling your own inferior one. – Mark Amery Jun 21 '16 at 22:57