After using cgi.parse_qs(), how to convert the result (dictionary) back to query string? Looking for something similar to urllib.urlencode().
4 Answers
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.
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 ofkey=valuepairs separated by'&'characters...
answered Oct 18, 2011 at 3:13
Ignacio Vazquez-Abrams
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9 Comments
johnsyweb
This is true, but the
dict returned by cgi.parse_qs() actually has lists as its "values". Passing these straight in will result in very odd looking query strings.johnsyweb
True enough. That'll teach me to read more than the first sentence!
Denilson Sá Maia
urllib.parse.urlencode in Python 3.
user1633272
The problem is that urlencode will convert space into +, which is not recommended.
Ignacio Vazquez-Abrams
@user1633272: Not recommended according to who?
|
In python3, slightly different:
from urllib.parse import urlencode
urlencode({'pram1': 'foo', 'param2': 'bar'})
output: 'pram1=foo¶m2=bar'
for python2 and python3 compatibility, try this:
try:
#python2
from urllib import urlencode
except ImportError:
#python3
from urllib.parse import urlencode
Comments
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)
2 Comments
Ignacio Vazquez-Abrams
Or you could pass
doseq=True.johnsyweb
@downvoter: How could I improve this answer? It has been six years since I wrote it!
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.
5 Comments
Ignacio Vazquez-Abrams
Have you met
str.join() yet? How about urllib.quote_plus()?
Anthony Cramp
@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.
garbados
Ah, indeed. Sorry for the cruddy implementation. Goes to show I should be more careful answering questions I've never faced myself.
harperville
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!
Mark Amery
@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.
cgi.parse_qs()is deprecated. Use urlparse.parse_qs() instead.