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<?xml-stylesheet href="/Style Library/st/xslt/rss2.xsl" type="text/xsl" media="screen" ?>
<rss version="2.0" xmlns:atom="http://www.w3.org/2005/Atom" xmlns:ta="http://www.smartraveller.gov.au/schema/rss/travel_advisories/" xmlns:dc="http://purl.org/dc/elements/1.1/">

  <channel>
    <title>Travel Advisories</title>
    <link>http://smartraveller.gov.au/countries/</link>
    <description>the Australian Department of Foreign Affairs and Trade's Smartraveller advisory service</description>
    <language>en</language>
    <webMaster>webmaster@dfat.gov.au</webMaster>
    <copyright>Copyright Commonwealth of Australia 2011</copyright>
    <ttl>60</ttl>
    <atom:link href="http://smartraveller.gov.au/countries/Documents/index.rss" rel="self" type="application/rss+xml" />
    <generator>zcms</generator>
    <image>
      <title>Advice</title>
      <link>http://smartraveller.gov.au/countries/</link>
      <url>/Style Library/st/images/dfat_logo_small.gif</url>
    </image>
    <item>
      <title>Czech Republic</title>
      <description>This travel advice has been reviewed. The level of our advice has not changed. Exercise normal safety precautions in the Czech Republic.</description>
      <link>http://smartraveller.gov.au/Countries/europe/eastern/Pages/czech_republic.aspx</link>
      <pubDate>26 Oct 2018 05:25:14 GMT</pubDate>
      <guid isPermaLink="false">cdbcc3d4-3a89-4768-ac1d-0221f8c99227 GMT</guid>
      <ta:warnings>
        <dc:coverage>Czech Republic</dc:coverage>
        <ta:level>2/5</ta:level>
        <dc:description>Exercise normal safety precautions</dc:description>
      </ta:warnings>
  </item>

I want to extract the value of <ta:level> under <warning> for each item I have. I alreay tried existing online solutions but nothing works for me. Basically, my xml contains multiple namespaces.

req = requests.request('GET', "https://smartraveller.gov.au/countries/documents/index.rss")
a = str(req.text).encode()
tree = etree.fromstring(a)

ns = {'TravelAd': 'https://smartraveller.gov.au/countries/documents/index.rss',
          'ta': 'http://www.smartraveller.gov.au/schema/rss/travel_advisories/'}

    e = tree.findall('{0}channel/{0}item/{0}warnings/{0}level'.format(ns))
    for i in e:
        print(i.text)
Ahtsham Manzoor
  • 23
  • 5

1 Answers1

0

The XML has multiple namespaces, but the only namespace you need to worry about is http://www.smartraveller.gov.au/schema/rss/travel_advisories/.

This is because the only elements in the path to your target that are in a namespace are ta:level and ta:warning.

Example...

from lxml import etree
import requests

req = requests.request('GET', "https://smartraveller.gov.au/countries/documents/index.rss")
a = str(req.text).encode()

tree = etree.fromstring(a)

ns = {'ta': 'http://www.smartraveller.gov.au/schema/rss/travel_advisories/'}

e = tree.findall('channel/item/ta:warnings/ta:level', ns)
for i in e:
    print(i.text)

prints...

2/5
2/5
4/5
2/5
...and so on

If you wanted a list, consider switching from findall() to xpath()...

e = tree.xpath('channel/item/ta:warnings/ta:level/text()', namespaces=ns)
print(e)

prints...

['2/5', '2/5', '4/5', '2/5', and so on...]
Daniel Haley
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  • 95