Question

I want to create a script in Python which downloads the current KML files of all the Maps I created on Google Maps.

To do so manually, I can use this:

http://maps.google.com.br/maps/ms?msid=USER_ID.MAP_ID&msa=0&output=kml

where USER_ID is a constant number Google uses to identify me, and MAP_ID is the individual map identifier generated by the link icon on top-right corner.

This is not very straightforward, because I have to manually browse “My Places” page on Google Maps, and get the links one by one.

From Google Maps API HTTP Protocol Reference:

The Map Feed is a feed of user-created maps.

This feed’s full GET URI is:

http://maps.google.com/maps/feeds/maps/default/full

This feed returns a list of all maps for the authenticated user.

** The page says this service is no longer available, so I wonder if there is a way to do the same in the present.

So, the question is: Is there a way to get/download the list of MAP_IDs of all my maps, preferrably using Python?

Thanks for reading

Answer 1

The correct answer to this question involves using Google Maps Data API, HTML interface, which by the way is deprecated but still solves my need in a more official way, or at least more convincing than parsing a web page. Here it goes:

# coding: utf-8

import urllib2, urllib, re, getpass

username = 'heltonbiker'
senha = getpass.getpass('Senha do usuário ' + username + ':')

dic = {
        'accountType':      'GOOGLE',
        'Email':            (username + '@gmail.com'),
        'Passwd':           senha,
        'service':          'local',
        'source':           'helton-mapper-1'
        }
url = 'https://www.google.com/accounts/ClientLogin?' + urllib.urlencode(dic)
output = urllib2.urlopen(url).read()
authid = output.strip().split('\n')[-1].split('=')[-1]

request = urllib2.Request('http://maps.google.com/maps/feeds/maps/default/full')
request.add_header('Authorization', 'GoogleLogin auth=%s' % authid)
source = urllib2.urlopen(request).read()

for link in re.findall('<link rel=.alternate. type=.text/html. href=((.)[^\1]*?)>', source):
    s = link[0]
    if 'msa=0' in s:
        print s

I arrived with this solution with a bunch of other questions in SO, and a lot of people helped me a lot, so I hope this code might help anyone else trying to do so in the future.