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Updated readme, and added MIT license
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coderholic committed Dec 1, 2010
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21 changes: 21 additions & 0 deletions LICENSE
@@ -0,0 +1,21 @@
The MIT License

Copyright (c) 2010 Ben Dowling

Permission is hereby granted, free of charge, to any person obtaining a copy
of this software and associated documentation files (the "Software"), to deal
in the Software without restriction, including without limitation the rights
to use, copy, modify, merge, publish, distribute, sublicense, and/or sell
copies of the Software, and to permit persons to whom the Software is
furnished to do so, subject to the following conditions:

The above copyright notice and this permission notice shall be included in
all copies or substantial portions of the Software.

THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR
IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY,
FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE
AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER
LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM,
OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN
THE SOFTWARE.
6 changes: 4 additions & 2 deletions README
Expand Up @@ -8,12 +8,14 @@ the way in which it gets cleaned up.

Includes 234 counties, 2,610 regions, 97,949 cities and 606 districts

For more information and examples see http://www.coderholic.com/django-cities-countries-regions-cities-and-districts-for-django/

Examples:
=========
Finding all London boroughs:

>>> london = City.objects.filter(country__name='United Kingdom').get(name='London')
>>> boroughs = Districts.objects.filter(city=london)
>>> boroughs = District.objects.filter(city=london)

Nearest city to a given lat,lon:

Expand All @@ -23,4 +25,4 @@ Nearest city to a given lat,lon:
5 Nearest cities to London:

>>> london = City.objects.filter(country__name='United Kingdom').get(name='London')
>>> nearest = City.objects.distance(london.location).exclude(id=london.id).orderby('distance')[:5]
>>> nearest = City.objects.distance(london.location).exclude(id=london.id).order_by('distance')[:5]

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