How to urlencode a querystring in Python?
PythonUrllibUrlencodePython Problem Overview
I am trying to urlencode this string before I submit.
queryString = 'eventName=' + evt.fields["eventName"] + '&' + 'eventDescription=' + evt.fields["eventDescription"];
Python Solutions
Solution 1 - Python
Python 2
What you're looking for is [urllib.quote_plus
][1]:
safe_string = urllib.quote_plus('string_of_characters_like_these:$#@=?%^Q^$')
#Value: '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
][2] (note the parse
child module)
import urllib.parse
safe_string = urllib.parse.quote_plus(...)
[1]: https://docs.python.org/2/library/urllib.html#urllib.quote_plus "urllib.quote_plus" [2]: https://docs.python.org/3/library/urllib.parse.html#urllib.parse.quote_plus
Solution 2 - Python
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
.
Solution 3 - Python
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.
Solution 4 - Python
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: https://stackoverflow.com/questions/526125)
- (see also: https://stackoverflow.com/questions/15479928)
- 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" : "[email protected]",
}
### ********************
## 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
"""
Solution 5 - Python
Solution 6 - Python
Try this:
urllib.pathname2url(stringToURLEncode)
urlencode
won't work because it only works on dictionaries. quote_plus
didn't produce the correct output.
Solution 7 - Python
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.
Solution 8 - Python
In Python 3, this worked with me
import urllib
urllib.parse.quote(query)
Solution 9 - Python
for future references (ex: for python3)
>>> import urllib.request as req
>>> query = 'eventName=theEvent&eventDescription=testDesc'
>>> req.pathname2url(query)
>>> 'eventName%3DtheEvent%26eventDescription%3DtestDesc'
Solution 10 - Python
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'
Solution 11 - Python
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" })
Solution 12 - Python
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
.
Solution 13 - Python
import urllib.parse
query = 'Hellö Wörld@Python'
urllib.parse.quote(query) // returns Hell%C3%B6%20W%C3%B6rld%40Python
Solution 14 - Python
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')