问题
I am trying to build a url so that I can send a get request to it using urllib
module.
Let's suppose my final_url
should be
url = "www.example.com/find.php?data=http%3A%2F%2Fwww.stackoverflow.com&search=Generate+value"
Now to achieve this I tried the following way:
>>> initial_url = "http://www.stackoverflow.com"
>>> search = "Generate+value"
>>> params = {"data":initial_url,"search":search}
>>> query_string = urllib.urlencode(params)
>>> query_string
'search=Generate%2Bvalue&data=http%3A%2F%2Fwww.stackoverflow.com'
Now if you compare my query_string
with the format of final_url
you can observer two things
1) The order of params are reversed instead of data=()&search=
it is search=()&data=
2) urlencode
also encoded the +
in Generate+value
I believe the first change is due to the random behaviour of dictionary. So, I though of using OrderedDict to reverse the dictionary. As, I am using python 2.6.5
I did
pip install ordereddict
But I am not able to use it in my code when I try
>>> od = OrderedDict((('a', 'first'), ('b', 'second')))
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
NameError: name 'OrderedDict' is not defined
So, my question is what is the correct way to use OrderedDict
in python 2.6.5 and how do I make urlencode
ignores the +
in Generate+value
.
Also, is this the correct approach to build URL
.
回答1:
You shouldn't worry about encoding the +
it should be restored on the server after unescaping the url. The order of named parameters shouldn't matter either.
Considering OrderedDict, it is not Python's built in. You should import it from collections
:
from urllib import urlencode, quote
# from urllib.parse import urlencode # python3
from collections import OrderedDict
initial_url = "http://www.stackoverflow.com"
search = "Generate+value"
query_string = urlencode(OrderedDict(data=initial_url,search=search))
url = 'www.example.com/find.php?' + query_string
if your python is too old and does not have OrderedDict in the module collections
, use:
encoded = "&".join( "%s=%s" % (key, quote(parameters[key], safe="+"))
for key in ordered(parameters.keys()))
Anyway, the order of parameters should not matter.
Note the safe
parameter of quote
. It prevents +
to be escaped, but it means , server will interpret Generate+value
as Generate value
. You can manually escape +
by writing %2B
and marking %
as safe char:
回答2:
First, the order of parameters in a http request should be completely irrelevant. If it isn't then the parsing library on the othe side is doing something wrong.
Second, of course the +
is encoded. +
is used as placeholder for a space in an encoded url, so if yor raw string contains a +
, this has to be escaped. urlencode
expects an unencoded string, you can't pass it a string that is already encoded.
回答3:
Some comments on the question and other answers:
- If you want to preserve order with
urllib.urlencode
, submit an ordered sequence of k/v pairs instead of mapping(dict). when you pass in a dict,urlencode
just callsfoo.items()
to grab an iterable sequence.
# urllib.urlencode accepts a mapping or sequence
# the output of this can vary, because `items()` is called on the dict
urllib.urlencode({"data": initial_url,"search": search})
# the output of this will not vary
urllib.urlencode((("data", initial_url), ("search", search)))
you can also pass in a secondard doseq
argument to adjust how iterable values are handled.
The order of parameters is not irrelevant. take these two urls for example:
https://example.com?foo=bar&bar=foo https://example.com?bar=foo&foo=bar
A http server should consider the order of these parameters irrelevant, but a function designed to compare URLs would not. In order to safely compare urls, these params would need to be sorted.
However, consider duplicate keys:
https://example.com?foo=3&foo=2&foo=1
The URI specs support duplicate keys, but don't address precedence or ordering.
In a given application, these could each trigger different results and be valid as well:
https://example.com?foo=1&foo=2&foo=3
https://example.com?foo=1&foo=3&foo=2
https://example.com?foo=2&foo=3&foo=1
https://example.com?foo=2&foo=1&foo=3
https://example.com?foo=3&foo=1&foo=2
https://example.com?foo=3&foo=2&foo=1
- The
+
is a reserved character that represents a space in a urlencoded form (vs%20
for part of the path).urllib.urlencode
escapes usingurllib.quote_plus()
, noturllib.quote()
. The OP most likely wanted to just do this:
initial_url = "http://www.stackoverflow.com"
search = "Generate value"
urllib.urlencode((("data", initial_url), ("search", search)))
Which produces:
data=http%3A%2F%2Fwww.stackoverflow.com&search=Generate+value
as the output.
来源:https://stackoverflow.com/questions/10765705/build-query-string-using-urlencode-python