如何在Python中使用Urlencode查询字符串?

问题:如何在Python中使用Urlencode查询字符串?

我尝试在提交之前对该字符串进行urlencode。

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

I am trying to urlencode this string before I submit.

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

回答 0

您需要将参数传递urlencode()为映射(dict)或2元组序列,例如:

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

Python 3或以上

采用:

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

请注意,这在通常意义上不会进行url编码(请看输出)。为此使用urllib.parse.quote_plus

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 1

Python 2

您正在寻找的是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

在Python 3中,该urllib软件包已分解为较小的组件。您将使用urllib.parse.quote_plus(注意parse子模块)

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

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 2

尝试使用请求而不是urllib,您无需费心urlencode!

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

编辑:

如果您需要有序的名称/值对或一个名称的多个值,请按如下所示设置参数:

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

而不是使用字典。

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 3

语境

  • Python(版本2.7.2)

问题

  • 您要生成一个urlencoded查询字符串。
  • 您有一个包含名称-值对的字典或对象。
  • 您希望能够控制名称-值对的输出顺序。

  • urllib.urlencode
  • urllib.quote_plus

陷阱

以下是一个完整的解决方案,包括如何处理一些陷阱。

### ********************
## 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
  """ 

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

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 4


回答 5

尝试这个:

urllib.pathname2url(stringToURLEncode)

urlencode将不起作用,因为它仅适用于词典。quote_plus没有产生正确的输出。

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 6

请注意,urllib.urlencode并非总能解决问题。问题在于某些服务关心参数的顺序,当您创建字典时,这些顺序会丢失。对于这种情况,如Ricky所建议的那样,urllib.quote_plus更好。

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

在Python 3中,这与我合作

import urllib

urllib.parse.quote(query)

In Python 3, this worked with me

import urllib

urllib.parse.quote(query)

回答 8

供将来参考(例如:适用于python3)

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

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 9

为了在需要同时支持python 2和python 3的脚本/程序中使用,这六个模块提供了quote和urlencode函数:

>>> 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'

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 10

如果urllib.parse.urlencode()给您错误,请尝试urllib3模块。

语法如下:

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

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 11

可能尚未提到的另一件事是,urllib.urlencode()它将字典中的空值编码为字符串,None而不是缺少该参数。我不知道通常是否需要这样做,但是不适合我的用例,因此我必须使用quote_plus

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.


回答 12

为使Python 3 urllib3正常工作,您可以根据其官方文档使用以下命令:

import urllib3

http = urllib3.PoolManager()
response = http.request(
     'GET',
     'https://api.prylabs.net/eth/v1alpha1/beacon/attestations',
     fields={  # here fields are the query params
          'epoch': 1234,
          'pageSize': pageSize 
      } 
 )
response = attestations.data.decode('UTF-8')

For Python 3 urllib3 works properly, you can use as follow as per its official docs :

import urllib3

http = urllib3.PoolManager()
response = http.request(
     'GET',
     'https://api.prylabs.net/eth/v1alpha1/beacon/attestations',
     fields={  # here fields are the query params
          'epoch': 1234,
          'pageSize': pageSize 
      } 
 )
response = attestations.data.decode('UTF-8')