问题

I am trying to urlencode this string before I submit.

queryString = 'eventName=' + evt.fields["eventName"] + '&' + 'eventDescription=' + evt.fields["eventDescription"];

回答1:

You need to pass your parameters into urlencode() as either a mapping (dict), or a sequence of 2-tuples, like:

>>> import urllib
>>> f = { 'eventName' : 'myEvent', 'eventDescription' : 'cool event'}
>>> urllib.urlencode(f)
'eventName=myEvent&eventDescription=cool+event'

Python 3 or above

Use:

>>> urllib.parse.urlencode(f)
eventName=myEvent&eventDescription=cool+event

Note that this does not do url encoding in the commonly used sense (look at the output). For that use urllib.parse.quote_plus.

回答2:

Python 2

What you're looking for is urllib.quote_plus:

>>> urllib.quote_plus('string_of_characters_like_these:$#@=?%^Q^$')
'string_of_characters_like_these%3A%24%23%40%3D%3F%25%5EQ%5E%24'

Python 3

In Python 3, the urllib package has been broken into smaller components. You'll use urllib.parse.quote_plus (note the parse child module)

import urllib.parse
urllib.parse.quote_plus(...)

回答3:

Try requests instead of urllib and you don't need to bother with urlencode!

import requests
requests.get('http://youraddress.com', params=evt.fields)

EDIT:

If you need ordered name-value pairs or multiple values for a name then set params like so:

params=[('name1','value11'), ('name1','value12'), ('name2','value21'), ...]

instead of using a dictionary.

回答4:

Context

Python (version 2.7.2 )

Problem

You want to generate a urlencoded query string.
You have a dictionary or object containing the name-value pairs.
You want to be able to control the output ordering of the name-value pairs.

Solution

urllib.urlencode
urllib.quote_plus

Pitfalls

dictionary output arbitrary ordering of name-value pairs
- (see also: Why is python ordering my dictionary like so?)
- (see also: Why is the order in dictionaries and sets arbitrary?)
handling cases when you DO NOT care about the ordering of the name-value pairs
handling cases when you DO care about the ordering of the name-value pairs
handling cases where a single name needs to appear more than once in the set of all name-value pairs

Example

The following is a complete solution, including how to deal with some pitfalls.

### ********************
## init python (version 2.7.2 )
import urllib

### ********************
## first setup a dictionary of name-value pairs
dict_name_value_pairs = {
  "bravo"   : "True != False",
  "alpha"   : "http://www.example.com",
  "charlie" : "hello world",
  "delta"   : "1234567 !@#$%^&*",
  "echo"    : "user@example.com",
  }

### ********************
## setup an exact ordering for the name-value pairs
ary_ordered_names = []
ary_ordered_names.append('alpha')
ary_ordered_names.append('bravo')
ary_ordered_names.append('charlie')
ary_ordered_names.append('delta')
ary_ordered_names.append('echo')

### ********************
## show the output results
if('NO we DO NOT care about the ordering of name-value pairs'):
  queryString  = urllib.urlencode(dict_name_value_pairs)
  print queryString 
  """
  echo=user%40example.com&bravo=True+%21%3D+False&delta=1234567+%21%40%23%24%25%5E%26%2A&charlie=hello+world&alpha=http%3A%2F%2Fwww.example.com
  """

if('YES we DO care about the ordering of name-value pairs'):
  queryString  = "&".join( [ item+'='+urllib.quote_plus(dict_name_value_pairs[item]) for item in ary_ordered_names ] )
  print queryString
  """
  alpha=http%3A%2F%2Fwww.example.com&bravo=True+%21%3D+False&charlie=hello+world&delta=1234567+%21%40%23%24%25%5E%26%2A&echo=user%40example.com
  """

回答5:

Python 3:

urllib.parse.quote_plus(string, safe='', encoding=None, errors=None)

回答6:

Note that the urllib.urlencode does not always do the trick. The problem is that some services care about the order of arguments, which gets lost when you create the dictionary. For such cases, urllib.quote_plus is better, as Ricky suggested.

回答7:

Try this:

urllib.pathname2url(stringToURLEncode)

urlencode won't work because it only works on dictionaries. quote_plus didn't produce the correct output.

回答8:

In Python 3, this worked with me

import urllib

urllib.parse.quote(query)

回答9:

for future references (ex: for python3)

>>> import urllib.request as req
>>> query = 'eventName=theEvent&eventDescription=testDesc'
>>> req.pathname2url(query)
>>> 'eventName%3DtheEvent%26eventDescription%3DtestDesc'

回答10:

For use in scripts/programs which need to support both python 2 and 3, the six module provides quote and urlencode functions:

>>> from six.moves.urllib.parse import urlencode, quote
>>> data = {'some': 'query', 'for': 'encoding'}
>>> urlencode(data)
'some=query&for=encoding'
>>> url = '/some/url/with spaces and %;!<>&'
>>> quote(url)
'/some/url/with%20spaces%20and%20%25%3B%21%3C%3E%26'

回答11:

If the urllib.parse.urlencode( ) is giving you errors , then Try the urllib3 module .

The syntax is as follows :

import urllib3
urllib3.request.urlencode({"user" : "john" })

回答12:

Another thing that might not have been mentioned already is that urllib.urlencode() will encode empty values in the dictionary as the string None instead of having that parameter as absent. I don't know if this is typically desired or not, but does not fit my use case, hence I have to use quote_plus.

来源：https://stackoverflow.com/questions/5607551/how-to-urlencode-a-querystring-in-python

标签

python

url-encoding

How to urlencode a querystring in Python?

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