Python ElementTree findall not working

Question

So i am trying to parse some opendata to build a database. Here's what I've done :

# -*- coding: utf-8 -*-
import urllib
import xml.etree.ElementTree as ET

url = 'http://opendata.cwb.gov.tw/govdownload?dataid=C-A0008-001&authorizationkey=rdec-key-123-45678-011121314'

root = ET.parse(urllib.urlopen(url)).getroot()

locations = root.findall('dataset/location')
print type(locations)
print "Counts:", len(locations)

it returned:

Counts: 0

I tried to parse some other xml data(change the url) and it worked fine

the xml data I'm working on is roughly like:

<?xml version="1.0" encoding="UTF-8"?><cwbopendata xmlns="urn:cwb:gov:tw:cwbcommon:0.1">
<identifier>0f819d32-297a-4512-9654-990a565bd080</identifier>
<sender>weather@cwb.gov.tw</sender>
<sent>2016-05-23T16:07:06+08:00</sent>
<status>Actual</status>
<msgType>Issue</msgType>
<dataid>CWB_A0008</dataid>
<scope>Public</scope>
<dataset>
    <location>
        <stationId>72C44</stationId>
        <time>
        <dataTime>105 4_2</dataTime>
        </time>
        <weatherElement>
        <elementName>平均氣溫</elementName>
        <elementValue>
        <value>21.1</value>
        </elementValue>
        .
        .
        .
    </location>
    <location>
    .
    .   
    .

Sorry I'm new to python and ElementTree and hope to get some good advices, thanks

score 11 · Answer 1 · edited May 23 '17 at 10:28

Your XML has default namespace which URI is 'urn:cwb:gov:tw:cwbcommon:0.1'. So all elements without prefix, within element where default namespace is declared would be considered in that namespace :

>>> ns = {'d': 'urn:cwb:gov:tw:cwbcommon:0.1'}
>>> locations = root.findall('d:dataset/d:location', ns)
>>> print "Counts:", len(locations)
Counts: 17

Related : Parsing XML with namespace in Python via 'ElementTree'

Python ElementTree findall not working

1 Answers1