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.

Flimm · Accepted Answer · 2017-01-05 13:37:26Z

78

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.

edited Jan 5, 2017 at 13:37

Flimm

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answered Jul 1, 2012 at 10:55

Qasim Khan

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5 Comments

Aaron Over a year ago

It's important to note that since QueryDict wraps an unordered dictionary, request.GET.urlencode() may not return keys in the same order that they appear in the original request URI.

uchuugaka Over a year ago

That's a fundamental of dictionaries actually. Additionally, most query strings should be considered unordered if they are keyed. If it is a true list it will be ordered.

cyberfly Over a year ago

For template usage, the syntax would be like this ` <a href="{{ request.path }}?{{ request.GET.urlencode }}">Link with all URL parameters</a> `

bparker Over a year ago

@uchuugaka This is no longer true since Python 3.7, insertion order is always remembered for dict keys now.

uchuugaka Over a year ago

mmm ok. I wrote that in 2016

aashah7 · Accepted Answer · 2017-10-22 05:09:06Z

66

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

Documentation for urllib.parse

edited Oct 22, 2017 at 5:09

aashah7

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answered Jul 1, 2012 at 9:49

Jon Clements

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3 Comments

Jon Clements Over a year ago

Okay, not quite sure why you'd want the query string and not have it processed, but using urlparse is the most readable and understandable way.

Alex Over a year ago

This really helped me - I needed a Map (or some object) that held the params from a request URL!

George Bezerra Over a year ago

This is very useful when you're not getting the URL from a request. In my case I had to parse pagination URLs that came from an external API response.

Flimm · Accepted Answer · 2017-01-05 13:38:37Z

60

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.

edited Jan 5, 2017 at 13:38

Flimm

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answered Mar 9, 2014 at 0:40

mynameistechno

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Comments

cyberfly · Accepted Answer · 2024-05-14 03:34:41Z

1

Additional information for the accepted answer

For template usage, the syntax would be something like this

<a href="{{ request.path }}?{{ request.GET.urlencode }}">Link with all URL parameters</a>

Alternative is using custom filter

https://stackoverflow.com/a/24135527/417899

answered May 14, 2024 at 3:34

cyberfly

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