If I have an object like:
d = {'a':1, 'en': 'hello'}
...then I can pass it to urllib.urlencode
, no problem:
percent_escaped = urlencode(d)
print percent_escaped
But if I try to pass an object with a value of type unicode
, game over:
d2 = {'a':1, 'en': 'hello', 'pt': u'olá'}
percent_escaped = urlencode(d2)
print percent_escaped # This fails with a UnicodeEncodingError
So my question is about a reliable way to prepare an object to be passed to urlencode
.
I came up with this function where I simply iterate through the object and encode values of type string or unicode:
def encode_object(object):
for k,v in object.items():
if type(v) in (str, unicode):
object[k] = v.encode('utf-8')
return object
This seems to work:
d2 = {'a':1, 'en': 'hello', 'pt': u'olá'}
percent_escaped = urlencode(encode_object(d2))
print percent_escaped
And that outputs a=1&en=hello&pt=%C3%B3la
, ready for passing to a POST call or whatever.
But my encode_object
function just looks really shaky to me. For one thing, it doesn't handle nested objects.
For another, I'm nervous about that if statement. Are there any other types that I should be taking into account?
And is comparing the type()
of something to the native object like this good practice?
type(v) in (str, unicode) # not so sure about this...
Thanks!
You should indeed be nervous. The whole idea that you might have a mixture of bytes and text in some data structure is horrifying. It violates the fundamental principle of working with string data: decode at input time, work exclusively in unicode, encode at output time.
Update in response to comment:
You are about to output some sort of HTTP request. This needs to be prepared as a byte string. The fact that urllib.urlencode is not capable of properly preparing that byte string if there are unicode characters with ordinal >= 128 in your dict is indeed unfortunate. If you have a mixture of byte strings and unicode strings in your dict, you need to be careful. Let's examine just what urlencode() does:
>>> import urllib
>>> tests = ['\x80', '\xe2\x82\xac', 1, '1', u'1', u'\x80', u'\u20ac']
>>> for test in tests:
... print repr(test), repr(urllib.urlencode({'a':test}))
...
'\x80' 'a=%80'
'\xe2\x82\xac' 'a=%E2%82%AC'
1 'a=1'
'1' 'a=1'
u'1' 'a=1'
u'\x80'
Traceback (most recent call last):
File "<stdin>", line 2, in <module>
File "C:\python27\lib\urllib.py", line 1282, in urlencode
v = quote_plus(str(v))
UnicodeEncodeError: 'ascii' codec can't encode character u'\x80' in position 0: ordinal not in range(128)
The last two tests demonstrate the problem with urlencode(). Now let's look at the str tests.
If you insist on having a mixture, then you should at the very least ensure that the str objects are encoded in UTF-8.
'\x80' is suspicious -- it is not the result of any_valid_unicode_string.encode('utf8').
'\xe2\x82\xac' is OK; it's the result of u'\u20ac'.encode('utf8').
'1' is OK -- all ASCII characters are OK on input to urlencode(), which will percent-encode such as '%' if necessary.
Here's a suggested converter function. It doesn't mutate the input dict as well as returning it (as yours does); it returns a new dict. It forces an exception if a value is a str object but is not a valid UTF-8 string. By the way, your concern about it not handling nested objects is a little misdirected -- your code works only with dicts, and the concept of nested dicts doesn't really fly.
def encoded_dict(in_dict):
out_dict = {}
for k, v in in_dict.iteritems():
if isinstance(v, unicode):
v = v.encode('utf8')
elif isinstance(v, str):
# Must be encoded in UTF-8
v.decode('utf8')
out_dict[k] = v
return out_dict
and here's the output, using the same tests in reverse order (because the nasty one is at the front this time):
>>> for test in tests[::-1]:
... print repr(test), repr(urllib.urlencode(encoded_dict({'a':test})))
...
u'\u20ac' 'a=%E2%82%AC'
u'\x80' 'a=%C2%80'
u'1' 'a=1'
'1' 'a=1'
1 'a=1'
'\xe2\x82\xac' 'a=%E2%82%AC'
'\x80'
Traceback (most recent call last):
File "<stdin>", line 2, in <module>
File "<stdin>", line 8, in encoded_dict
File "C:\python27\lib\encodings\utf_8.py", line 16, in decode
return codecs.utf_8_decode(input, errors, True)
UnicodeDecodeError: 'utf8' codec can't decode byte 0x80 in position 0: invalid start byte
>>>
Does that help?