How can I make unique URL in Python a la http://imgur.com/gM19g or http://tumblr.com/xzh3bi25y When using uuid from python I get a very large one. I want something shorter f
Python's short_url is awesome.
Here is an example:
import short_url
id = 20 # your object id
domain = 'mytiny.domain'
shortened_url = "http://{}/{}".format(
domain,
short_url.encode_url(id)
)
And to decode the code:
decoded_id = short_url.decode_url(param)
That's it :)
Hope this will help.
This answer comes pretty late but I stumbled upon this question when I was planning to create an URL shortener project. Now that I have implemented a fully functional URL shortener(source code at amitt001/pygmy) I am adding an answer here for others.
The basic principle behind any URL shortener is to get an int from long URL then use base62(base32, etc) encoding to convert this int to a more readable short URL.
How is this int generated?
Most of the URL shortener uses some auto-incrementing datastore to add URL to datastore and use the autoincrement id to get base62 encoding of int.
The sample base62 encoding from string program:
# Base-62 hash
import string
import time
_BASE = 62
class HashDigest:
"""Base base 62 hash library."""
def __init__(self):
self.base = string.ascii_letters + string.digits
self.short_str = ''
def encode(self, j):
"""Returns the repeated div mod of the number.
:param j: int
:return: list
"""
if j == 0:
return [j]
r = []
dividend = j
while dividend > 0:
dividend, remainder = divmod(dividend, _BASE)
r.append(remainder)
r = list(reversed(r))
return r
def shorten(self, i):
"""
:param i:
:return: str
"""
self.short_str = ""
encoded_list = self.encode(i)
for val in encoded_list:
self.short_str += self.base[val]
return self.short_str
This is just the partial code showing base62 encoding. Check out the complete base62 encoding/decoding code at core/hashdigest.py
All the link in this answer are shortened from the project I created
It doesn't really matter that this is Python, but you just need a hash function that maps to the length you want. For example, maybe use MD5 and then take just the first n
characters. You'll have to watch out for collisions in that case, though, so you might want to pick something a little more robust in terms of collision detection (like using primes to cycle through the space of hash strings).
I don't know if you can use this, but we generate content objects in Zope that get unique numeric ids based on current time strings, in millis (eg, 1254298969501)
Maybe you can guess the rest. Using the recipe described here: How to convert an integer to the shortest url-safe string in Python?, we encode and decode the real id on the fly, with no need for storage. A 13-digit integer is reduced to 7 alphanumeric chars in base 62, for example.
To complete the implementation, we registered a short (xxx.yy) domain name, that decodes and does a 301 redirect for "not found" URLs,
If I was starting over, I would subtract the "starting-over" time (in millis) from the numeric id prior to encoding, then re-add it when decoding. Or else when generating the objects. Whatever. That would be way shorter..
I'm not sure most URL shorteners use a random string. My impression is they write the URL to a database, then use the integer ID of the new record as the short URL, encoded base 36 or 62 (letters+digits).
Python code to convert an int to a string in arbitrary bases is here.
Edit: Here, I wrote a module for you. Use it. http://code.activestate.com/recipes/576918/
Counting up from 1 will guarantee short, unique URLS. /1, /2, /3 ... etc.
Adding uppercase and lowercase letters to your alphabet will give URLs like those in your question. And you're just counting in base-62 instead of base-10.
Now the only problem is that the URLs come consecutively. To fix that, read my answer to this question here:
Map incrementing integer range to six-digit base 26 max, but unpredictably
Basically the approach is to simply swap bits around in the incrementing value to give the appearance of randomness while maintaining determinism and guaranteeing that you don't have any collisions.