URL encoding in python

2020-02-02 07:57发布

Is there a simple method I'm missing in urllib or other library for this task? URL encoding replaces unsafe ASCII characters with a "%" followed by two hexadecimal digits.

Here's an example of an input and my expected output:

Mozilla/5.0 (Linux; U; Android 4.0; xx-xx; Galaxy Nexus Build/IFL10C) AppleWebKit/534.30 (KHTML, like Gecko) Version/4.0 Mobile Safari/534.30

Mozilla%2F5.0+%28Linux%3B+U%3B+Android+4.0%3B+xx-xx%3B+Galaxy+Nexus+Build%2FIFL10C%29+AppleWebKit%2F534.30+%28KHTML%2C+like+Gecko%29+Version%2F4.0+Mobile+Safari%2F534.30

3条回答
一纸荒年 Trace。
2楼-- · 2020-02-02 08:34

For Python 2.x, use urllib.quote

Replace special characters in string using the %xx escape. Letters, digits, and the characters '_.-' are never quoted. By default, this function is intended for quoting the path section of the URL. The optional safe parameter specifies additional characters that should not be quoted — its default value is '/'.

example:

In [1]: import urllib

In [2]: urllib.quote('%')
Out[2]: '%25'

EDIT:

In your case, in order to replace space by plus signs, you may use urllib.quote_plus

example:

In [4]: urllib.quote_plus('a b')
Out[4]: 'a+b'

For Python 3.x, use quote

>>> import urllib
>>> a = "asdas#@das"
>>> urllib.parse.quote(a)
'asdas%23%40das'

and for string with space use quote_plus

>>> import urllib
>>> a = "as da& s#@das"
>>> urllib.parse.quote_plus(a)
'as+da%26+s%23%40das'
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劳资没心,怎么记你
3楼-- · 2020-02-02 08:51

Keep in mind that both urllib.quote and urllib.quote_plus throw an error if an input is a unicode string:

s = u'\u2013'
urllib.quote(s)

Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "C:\Python27\lib\urllib.py", line 1303, in quote
    return ''.join(map(quoter, s))
KeyError: u'\u2013'

As answered here on SO, one has to use 'UTF-8' explicitly:

urllib.quote(s.encode('utf-8'))
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Bombasti
4楼-- · 2020-02-02 08:55

Also, if you have a dict of several values, the best way to do it will be urllib.urlencode.

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