blocks|key|4832604|text|像这样的东西？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4832605|a+=+{'a':+'b',+'c':+{'d':+'e'}}

url+=+urllib.urlencode([('%25s[%25s]'%25(k,v.keys()[0]),+v.values()[0]+)+if+type(v)==dict+else+(k,v)+for+k,v+in+a.iteritems()])

url+=+'a=b&c%255Bd%255D=e'|code-block|syntax|javascript|4832606|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Something like this?

<pre><code>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&amp;c%5Bd%5D=e'
</code></pre>

blocks|key|1229943|text|那json.dumps和json.loads呢？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1229944|d+=+{'a':+'b',+'c':+{'d':+'e'}}
s+=+json.dumps(d)++#+s:+'{"a":+"b",+"c":+{"d":+"e"}}'
json.loads(s)++#+->+d|code-block|syntax|javascript|1229945|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

what about json.dumps and json.loads?

<pre><code>d = {'a': 'b', 'c': {'d': 'e'}}
s = json.dumps(d) # s: '{"a": "b", "c": {"d": "e"}}'
json.loads(s) # -&gt; d
</code></pre>

blocks|key|1977028|text|好了，各位。我自己实现的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1977029|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%2526c&j=k&d[e][f%2526g]=h%252Ai'
++++"""
++++def+recursion(d,+base=[]):
++++++++pairs+=+[]

++++++++for+key,+value+in+d.items():
++++++++++++new_base+=+base+%2B+[key]
++++++++++++if+hasattr(value,+'values'):
++++++++++++++++pairs+%2B=+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+=+"%25s[%25s]=%25s"+%25+(first,+']['.join(rest),+urllib.quote(unicode(value)))
++++++++++++++++else:
++++++++++++++++++++new_pair+=+"%25s=%25s"+%25+(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()|code-block|syntax|javascript|1977030|不过，我还是很有兴趣知道是否有更好的方法来做这件事。我不敢相信Python的标准库没有实现这一点。|1977031|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

OK people. I implemented it myself:

<pre><code>import urllib

def recursive_urlencode(d):
 """URL-encode a multidimensional dictionary.

 &gt;&gt;&gt; data = {'a': 'b&amp;c', 'd': {'e': {'f&amp;g': 'h*i'}}, 'j': 'k'}
 &gt;&gt;&gt; recursive_urlencode(data)
 u'a=b%26c&amp;j=k&amp;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) &gt; 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 '&amp;'.join(recursion(d))

if __name__ == "__main__":
 import doctest
 doctest.testmod()
</code></pre>

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.

blocks|key|3421147|text|这个简化的版本怎么样：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3421148|def+_clean(value):
++++return+urllib.quote(unicode(value))

'&'.join([+v+for+val+in+[[+"%25s[%25s]=%25s"%25(k,ik,+_(iv))+
++++for+ik,+iv+in+v.items()]+if+type(v)==dict+else+["%25s=%25s"%25(k,_(v))]+
++++for+k,v+in+data.items()+]+
++++for+v+in+val+])|code-block|syntax|javascript|3421149|我同意这是不可读的，也许扁平化列表可以用itertools.chain做得更好，而不是另一个列表理解。|3421150|这只深入了一个级别，如果你想添加一些逻辑来管理N个"%25s“，那么你可以再深入N个级别，但我猜这并不是必要的|3421151|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

what about this simplified version:

<pre><code>def _clean(value):
 return urllib.quote(unicode(value))

'&amp;'.join([ v for val in [[ "%s[%s]=%s"%(k,ik, _(iv)) 
 for ik, iv in v.items()] if type(v)==dict else ["%s=%s"%(k,_(v))] 
 for k,v in data.items() ] 
 for v in val ])
</code></pre>

I agree is not readable, maybe flattening the list can be better done with itertools.chain instead of another list comprehension.

This only goes 1 level deeper, yours can go N levels deeper if you would add some logic to manage N numbers of "[%s]" depending on the level, but I guess is not that necesary

blocks|key|3421242|text|上面的解决方案只适用于深度小于2的数组。下面的代码将正确地对任意深度的多维数组进行urlencode。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3421243|#!/usr/bin/env+python

import+sys
import+urllib

def+recursive_urlencode(data):
++++def+r_urlencode(data,+parent=None,+pairs=None):
++++++++if+pairs+is+None:
++++++++++++pairs+=+{}
++++++++if+parent+is+None:
++++++++++++parents+=+[]
++++++++else:
++++++++++++parents+=+parent

++++++++for+key,+value+in+data.items():
++++++++++++if+hasattr(value,+'values'):
++++++++++++++++parents.append(key)
++++++++++++++++r_urlencode(value,+parents,+pairs)
++++++++++++++++parents.pop()
++++++++++++else:
++++++++++++++++pairs[renderKey(parents+%2B+[key])]+=+renderVal(value)

++++++++return+pairs
++++return+urllib.urlencode(r_urlencode(data))


def+renderKey(parents):
++++depth,+outStr+=+0,+''
++++for+x+in+parents:
++++++++str+=+"[%25s]"+if+depth+>+0+else+"%25s"
++++++++outStr+%2B=+str+%25+renderVal(x)
++++++++depth+%2B=+1
++++return+outStr


def+renderVal(val):
++++return+urllib.quote(unicode(val))


def+main():
++++print+recursive_urlencode(payload)


if+__name__+==+'__main__':
++++sys.exit(main())|code-block|syntax|javascript|3421244|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

The above solution only works for arrays with depth &lt; 2. The code below will properly urlencode a multidimensional array of any depth.

<pre><code>#!/usr/bin/env python

import sys
import urllib

def recursive_urlencode(data):
 def r_urlencode(data, parent=None, pairs=None):
 if pairs is None:
 pairs = {}
 if parent is None:
 parents = []
 else:
 parents = parent

 for key, value in data.items():
 if hasattr(value, 'values'):
 parents.append(key)
 r_urlencode(value, parents, pairs)
 parents.pop()
 else:
 pairs[renderKey(parents + [key])] = renderVal(value)

 return pairs
 return urllib.urlencode(r_urlencode(data))


def renderKey(parents):
 depth, outStr = 0, ''
 for x in parents:
 str = "[%s]" if depth &gt; 0 else "%s"
 outStr += str % renderVal(x)
 depth += 1
 return outStr


def renderVal(val):
 return urllib.quote(unicode(val))


def main():
 print recursive_urlencode(payload)


if __name__ == '__main__':
 sys.exit(main())
</code></pre>

blocks|key|1977174|text|基于@malaney的代码，我认为下面的代码很好地模拟了PHP函数http_build_query()。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1977175|#!/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+=+"[%25s]"+if+depth+>+0+or+isinstance(x,+int)+else+"%25s"
++++++++++++outStr+%2B=+s+%25+str(x)
++++++++++++depth+%2B=+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&ampersand'
++++}
++++print(http_build_query(payload))

++++payload2+=+[
++++++++'item1',
++++++++'item2'
++++]
++++print(http_build_query(payload2))|code-block|syntax|javascript|1977176|entityMap^0|X|I|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

Based on the code of @malaney, I think that the code below emulates the PHP function <code>http_build_query()</code> quite well.

<pre><code>#!/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 &gt; 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&amp;ampersand'
 }
 print(http_build_query(payload))

 payload2 = [
 'item1',
 'item2'
 ]
 print(http_build_query(payload2))
</code></pre>

blocks|key|1972459|text|如果你想把python+dict/list/nested转换成PHP+Array，就像urlencoded字符串一样。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1972460|在python中，您想要转换为urlencoded的大多数数据类型可能是：dict+list+tuple+nested+of+them，如|offset|length|style|CODE|1972461|a+=+[1,+2]
print(recursive_urlencode(a))
#+0=1&1=2


a2+=+(1,+'2')
print(recursive_urlencode(a2))
#+0=1&1=2


b+=+{'a':+11,+'b':+'foo'}
print(recursive_urlencode(b))
#+a=11&b=foo


c+=+{'a':+11,+'b':+[1,+2]}
print(recursive_urlencode(c))
#+a=11&b[0]=1&b[1]=2


d+=+[1,+{'a':+11,+'b':+22}]
print(recursive_urlencode(d))
#+0=1&1[a]=11&1[b]=22


e+=+{'a':+11,+'b':+[1,+{'c':+123},+[3,+'foo']]}
print(recursive_urlencode(e))
#+a=11&b[0]=1&b[1][c]=123&b[2][0]=3&b[2][1]=foo|code-block|syntax|javascript|1972462|https://github.com/Viky-zhang/to_php_post_arr|1972463|附注：https://stackoverflow.com/a/4014164/2752670中的一些代码|1972464|entityMap|0|LINK|mutability|MUTABLE|url|1|https://stackoverflow.com/a/4014164/2752670^0|0|11|4|16|4|1B|5|1H|E|0|0|0|19|0|0|3|17|1|0^^$0|@$1|2|3|4|5|6|7|Z|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|10|8|@$D|11|E|12|F|G]|$D|13|E|14|F|G]|$D|15|E|16|F|G]|$D|17|E|18|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|19|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|1A|8|@]|9|@$D|1B|E|1C|1|1D]]|A|$]]|$1|O|3|P|5|6|7|1E|8|@]|9|@$D|1F|E|1G|1|1H]]|A|$]]|$1|Q|3|-4|5|6|7|1I|8|@]|9|@]|A|$]]]|R|$S|$5|T|U|V|A|$W|N]]|X|$5|T|U|V|A|$W|Y]]]]

If you want to convert python dict/list/nested to PHP Array like urlencoded string.

In python, most of the data type you want to convert to urlencoded maybe: <code>dict</code> <code>list</code> <code>tuple</code> <code>nested of them</code>, Like

<pre><code>a = [1, 2]
print(recursive_urlencode(a))
# 0=1&amp;1=2


a2 = (1, '2')
print(recursive_urlencode(a2))
# 0=1&amp;1=2


b = {'a': 11, 'b': 'foo'}
print(recursive_urlencode(b))
# a=11&amp;b=foo


c = {'a': 11, 'b': [1, 2]}
print(recursive_urlencode(c))
# a=11&amp;b[0]=1&amp;b[1]=2


d = [1, {'a': 11, 'b': 22}]
print(recursive_urlencode(d))
# 0=1&amp;1[a]=11&amp;1[b]=22


e = {'a': 11, 'b': [1, {'c': 123}, [3, 'foo']]}
print(recursive_urlencode(e))
# a=11&amp;b[0]=1&amp;b[1][c]=123&amp;b[2][0]=3&amp;b[2][1]=foo
</code></pre>

<a href="https://github.com/Viky-zhang/to_php_post_arr" rel="nofollow noreferrer">https://github.com/Viky-zhang/to_php_post_arr</a>

P.S. some code from: <a href="https://stackoverflow.com/a/4014164/2752670">https://stackoverflow.com/a/4014164/2752670</a>

blocks|key|1227611|text|函数get_encoded_url_params()接受dict作为参数，并返回dict的url编码形式。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1227612|def+get_encoded_url_params(d):
++++"""URL-encode+a+nested+dictionary.
++++:param+d+=+dict
++++:returns+url+encoded+string+with+dict+key-value+pairs+as+query+parameters
++++e.g.
++++if+d+=+{+"addr":{+"country":+"US",+"line":+["a","b"]+},
++++++++++++"routing_number":+"011100915",+"token":+"asdf"
++++++++}
++++:returns+'addr[country]=US&addr[line][0]=a&addr[line][1]=b&routing_number=011100915&token=asdf'
++++or+'addr%255Bcountry%255D=US&addr%255Bline%255D%255B0%255D=a&addr%255Bline%255D%255B1%255D=b&routing_number=011100915&token=asdf'
++++(which+is+url+encoded+form+of+the+former+using+quote_plus())
++++"""

++++def+get_pairs(value,+base):
++++++++if+isinstance(value,+dict):
++++++++++++return+get_dict_pairs(value,+base)
++++++++elif+isinstance(value,+list):
++++++++++++return+get_list_pairs(value,+base)
++++++++else:
++++++++++++return+[base+%2B+'='+%2B+str(value)]
++++++++++++#+use+quote_plus()+to+get+url+encoded+string
++++++++++++#+return+[quote_plus(base)+%2B+'='+%2B+quote_plus(str(value))]

++++def+get_list_pairs(li,+base):
++++++++pairs+=+[]
++++++++for+idx,+value+in+enumerate(li):
++++++++++++new_base+=+base+%2B+'['+%2B+str(idx)+%2B+']'
++++++++++++pairs+%2B=+get_pairs(value,+new_base)
++++++++return+pairs

++++def+get_dict_pairs(d,+base=''):
++++++++pairs+=+[]
++++++++for+key,+value+in+d.items():
++++++++++++new_base+=+key+if+base+==+''+else+base+%2B+'['+%2B+key+%2B+']'
++++++++++++pairs+%2B=+get_pairs(value,+new_base)
++++++++return+pairs

++++return+'&'.join(get_dict_pairs(d))|code-block|syntax|javascript|1227613|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

The function get_encoded_url_params() takes a dict as argument and returns url encoded form of the dict.

<pre><code>def get_encoded_url_params(d):
 """URL-encode a nested dictionary.
 :param d = dict
 :returns url encoded string with dict key-value pairs as query parameters
 e.g.
 if d = { "addr":{ "country": "US", "line": ["a","b"] },
 "routing_number": "011100915", "token": "asdf"
 }
 :returns 'addr[country]=US&amp;addr[line][0]=a&amp;addr[line][1]=b&amp;routing_number=011100915&amp;token=asdf'
 or 'addr%5Bcountry%5D=US&amp;addr%5Bline%5D%5B0%5D=a&amp;addr%5Bline%5D%5B1%5D=b&amp;routing_number=011100915&amp;token=asdf'
 (which is url encoded form of the former using quote_plus())
 """

 def get_pairs(value, base):
 if isinstance(value, dict):
 return get_dict_pairs(value, base)
 elif isinstance(value, list):
 return get_list_pairs(value, base)
 else:
 return [base + '=' + str(value)]
 # use quote_plus() to get url encoded string
 # return [quote_plus(base) + '=' + quote_plus(str(value))]

 def get_list_pairs(li, base):
 pairs = []
 for idx, value in enumerate(li):
 new_base = base + '[' + str(idx) + ']'
 pairs += get_pairs(value, new_base)
 return pairs

 def get_dict_pairs(d, base=''):
 pairs = []
 for key, value in d.items():
 new_base = key if base == '' else base + '[' + key + ']'
 pairs += get_pairs(value, new_base)
 return pairs

 return '&amp;'.join(get_dict_pairs(d))
</code></pre>

How can I get a URL-encoded version of a multidimensional dictionary in Python? Unfortunately, <code>urllib.urlencode()</code> only works in a single dimension. I would need a version capable of recursively encoding the dictionary.

For example, if I have the following dictionary:

<pre><code>{'a': 'b', 'c': {'d': 'e'}}
</code></pre>

I want to obtain the following string:

<pre><code>a=b&amp;c[d]=e
</code></pre>

urlencode a multidimensional dictionary in python

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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如何在Python中获得多维字典的URL编码版本？不幸的是，urllib.urlencode()只能在一个维度上工作。我需要一个能够递归编码字典的版本。例如，如果我有以下字典：{'a': 'b', 'c': {'d': 'e'}}我想要获取以下字符串：a=b&c[d]=e

问用python对多维字典进行urlencode
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