urlencode a multidimensional dictionary in python
Based on the code of @malaney, I think that the code below emulates the PHP function http_build_query()
quite well.
#!/usr/bin/env python3
import urllib.parse
def http_build_query(data):
parents = list()
pairs = dict()
def renderKey(parents):
depth, outStr = 0, ''
for x in parents:
s = "[%s]" if depth > 0 or isinstance(x, int) else "%s"
outStr += s % str(x)
depth += 1
return outStr
def r_urlencode(data):
if isinstance(data, list) or isinstance(data, tuple):
for i in range(len(data)):
parents.append(i)
r_urlencode(data[i])
parents.pop()
elif isinstance(data, dict):
for key, value in data.items():
parents.append(key)
r_urlencode(value)
parents.pop()
else:
pairs[renderKey(parents)] = str(data)
return pairs
return urllib.parse.urlencode(r_urlencode(data))
if __name__ == '__main__':
payload = {
'action': 'add',
'controller': 'invoice',
'code': 'debtor',
'InvoiceLines': [
{'PriceExcl': 150, 'Description': 'Setupfee'},
{'PriceExcl':49.99, 'Description':'Subscription'}
],
'date': '2016-08-01',
'key': 'Yikes&ersand'
}
print(http_build_query(payload))
payload2 = [
'item1',
'item2'
]
print(http_build_query(payload2))
Something like this?
a = {'a': 'b', 'c': {'d': 'e'}}
url = urllib.urlencode([('%s[%s]'%(k,v.keys()[0]), v.values()[0] ) if type(v)==dict else (k,v) for k,v in a.iteritems()])
url = 'a=b&c%5Bd%5D=e'
OK people. I implemented it myself:
import urllib
def recursive_urlencode(d):
"""URL-encode a multidimensional dictionary.
>>> data = {'a': 'b&c', 'd': {'e': {'f&g': 'h*i'}}, 'j': 'k'}
>>> recursive_urlencode(data)
u'a=b%26c&j=k&d[e][f%26g]=h%2Ai'
"""
def recursion(d, base=[]):
pairs = []
for key, value in d.items():
new_base = base + [key]
if hasattr(value, 'values'):
pairs += recursion(value, new_base)
else:
new_pair = None
if len(new_base) > 1:
first = urllib.quote(new_base.pop(0))
rest = map(lambda x: urllib.quote(x), new_base)
new_pair = "%s[%s]=%s" % (first, ']['.join(rest), urllib.quote(unicode(value)))
else:
new_pair = "%s=%s" % (urllib.quote(unicode(key)), urllib.quote(unicode(value)))
pairs.append(new_pair)
return pairs
return '&'.join(recursion(d))
if __name__ == "__main__":
import doctest
doctest.testmod()
Still, I'd be interested to know if there's a better way to do this. I can't believe Python's standard library doesn't implement this.