Best way to get query string from a URL in python?
Question:
I need to get the query string from this URL https://stackoverflow.com/questions/ask?next=1&value=3 and I don’t want to use request.META. I have figured out that there are two more ways to get the query string:
-
Using urlparse
urlparse.urlparse(url).query
-
Using url encode
Use urlencode and pass the request.GET params dictionary into it to get the string representation.
So which way is better? My colleagues prefer urlencode but have not provided a satisfying explanation. They claim that urlparse calls urlencode internally which is something I’m not sure about since urlencode lives in the urllib module.
Answers:
Third option:
>>> from urlparse import urlparse, parse_qs
>>> url = 'http://something.com?blah=1&x=2'
>>> urlparse(url).query
'blah=1&x=2'
>>> parse_qs(urlparse(url).query)
{'blah': ['1'], 'x': ['2']}
In Python 3+ this is available as:
from urllib.parse import parse_qs
You can make Query string using GET parameters like this
request.GET.urlencode()
This does not include the ?
prefix, and it may not return the keys in the same order as in the original request.
I prefer using
request.META['QUERY_STRING']
From docs:
https://docs.djangoproject.com/en/stable/ref/request-response/#django.http.HttpRequest.META
This does not include the ?
prefix.
I need to get the query string from this URL https://stackoverflow.com/questions/ask?next=1&value=3 and I don’t want to use request.META. I have figured out that there are two more ways to get the query string:
-
Using urlparse
urlparse.urlparse(url).query -
Using url encode
Use urlencode and pass the request.GET params dictionary into it to get the string representation.
So which way is better? My colleagues prefer urlencode but have not provided a satisfying explanation. They claim that urlparse calls urlencode internally which is something I’m not sure about since urlencode lives in the urllib module.
Third option:
>>> from urlparse import urlparse, parse_qs
>>> url = 'http://something.com?blah=1&x=2'
>>> urlparse(url).query
'blah=1&x=2'
>>> parse_qs(urlparse(url).query)
{'blah': ['1'], 'x': ['2']}
In Python 3+ this is available as:
from urllib.parse import parse_qs
You can make Query string using GET parameters like this
request.GET.urlencode()
This does not include the ?
prefix, and it may not return the keys in the same order as in the original request.
I prefer using
request.META['QUERY_STRING']
From docs:
https://docs.djangoproject.com/en/stable/ref/request-response/#django.http.HttpRequest.META
This does not include the ?
prefix.