Python XML parse

Question

I am trying to parse an XML file retrieved from OCTranspo (Ottawa City Bus Company) using Python. My problem is that I can't seem to access the sub-fields, such as Latitude and Longitude.

Here is a heavily shortened version of a sample xml file, that still results in the problem:

<?xml version="1.0" encoding="utf-8"?>
<soap:Envelope xmlns:soap="http://schemas.xmlsoap.org/soap/envelope/"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns:xsd="http://www.w3.org/2001/XMLSchema">
<soap:Body>

<Route xmlns="http://tempuri.org/">

<Trips>
<Trip><TripDestination>Barrhaven Centre</TripDestination
<TripStartTime>19:32</TripStartTime><Latitude>45.285458</Latitude
<Longitude>-75.746786</Longitude></Trip>
</Trips>

</Route>

</soap:Body>
</soap:Envelope>

And here's my code:

import xml.etree.ElementTree as ET
import urllib

u = urllib.urlopen('https://api.octranspo1.com/v1.1/GetNextTripsForStop', 'appID=7a51d100&apiKey=5c5a8438efc643286006d82071852789&routeNo=95&stopNo=3044')
data = u.read()

f = open('route3044.xml', 'wb')
f.write(data)
f.close()

doc = ET.parse('route3044.xml')

for bus in doc.findall('Trip'):
    lat = bus.findtext('Latitude')
    #NEVER EXECUTES
    print trip

If I execute the same code against a very simple xml file (one without the soap:Envelope...) then the code works flawlessly. However, as the xml I need is generated by OCTranspo I can't control the format.

I'm not sure if the issue is a 'namespace' issue or a bug in Python.

Any assistance would be appreciated.

UPDATE: 21-Sept-2013

I changed the code that searches for the Lat and Lon to this:

doc = ET.parse('Stop1A.xml')

for a in doc.findall('{http://schemas.xmlsoap.org/soap/envelope/}Body'):
    for b in a.findall('{http://octranspo.com}GetNextTripsForStopResponse'): 
        for c in b.findall('{http://octranspo.com}GetNextTripsForStopResult'):   
            for d in c.findall('{http://tempuri.org/}Route'):
                for e in d.findall('{http://tempuri.org/}RouteDirection'):
                    direction = e.findtext('{http://tempuri.org/}Direction')
                    if direction == 'Eastbound':
                        for f in e.findall('{http://tempuri.org/}Trips'):
                            for g in f.findall('{http://tempuri.org/}Trip'):
                                lat = g.findtext('{http://tempuri.org/}Latitude')
                                lon = g.findtext('{http://tempuri.org/}Longitude')
                                print lat + ',' + lon
                                print 'Done'

End result is that I now can see the 'Eastbound' buses on route 95. I know this code is not pretty, but it works. My next goal will be to optimize with perhaps using namespace tricks.

If anyone cares to try accessing the url, note that it's common to see 'no buses' for 5-7 mins, as the url simply returns the closest 6 buses to the stop. Three buses going Eastbound, and three buses going Westbound. If the closest bus is over 7 mins away then the return is null. The code returns the Lat and Lon of a bus - which I can then plot the location using Google Maps.

Kelly

rici · Accepted Answer · 2013-09-21 02:13:49Z

2

According to the ElementTree documentation:

Element.findall() finds only elements with a tag which are direct children of the current element. (emphasis added)

Fortunately, ElementTree has XPath support

Change doc.findall('Trip') (search through direct children of doc) to doc.findall('.//Trip') (search recursively for a child of doc), and it should work as you expected.

answered Sep 21, 2013 at 2:13

rici

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2 Comments

Scooter Over a year ago

Thanks for the response. Unfortunately the change did not work.

Scooter Over a year ago

I stand corrected. I didn't append the namespace to the search so this does work. Thanks!

mzjn · Accepted Answer · 2013-09-21 13:06:26Z

1

Here is a simple way to get latitudes and longitudes for each trip. You don't need to iterate through every element. Note the use of .// to find all {http://tempuri.org/}Trip elements.

import xml.etree.ElementTree as ET

doc = ET.parse("temp.xml")     # Your shortened XML document

for bus in doc.findall('.//{http://tempuri.org/}Trip'):
    lat = bus.findtext('{http://tempuri.org/}Latitude')
    lon = bus.findtext('{http://tempuri.org/}Longitude')
    print lat, lon

Output:

45.285458 -75.746786

answered Sep 21, 2013 at 13:06

mzjn

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1 Comment

Scooter Over a year ago

Much more elegant. Thanks. I've just recently switched over to Python and find it quite nice to work with.

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