Python namespaced xml-parsing error

Question

What's wrong with the following code? I am expecting boogie as output.

import urllib.request
from html.parser import HTMLParser
import xml.etree.ElementTree as ET

html = '''<rdf:RDF xmlns:rdf="http://www.w3.org/1999/02/22-rdf-syntax-ns#"
 xmlns="http://purl.org/rss/1.0/"
 xmlns:enc="http://purl.oclc.org/net/rss_2.0/enc#"
><foo><title>boogie</title></foo></rdf:RDF>'''

root = ET.fromstring(html)
ns = { 'default': 'http://purl.org/rss/1.0/', 'rdf': 'http://www.w3.org/1999/02/22-rdf-syntax-ns#'}

titles = root.findall("default:.//title", ns)
[print(title.text) for title in titles]

gaback · Accepted Answer · 2017-07-07 23:12:35Z

1

import urllib.request
from html.parser import HTMLParser
import xml.etree.ElementTree as ET

html = '''<rdf:RDF xmlns:rdf="http://www.w3.org/1999/02/22-rdf-syntax-ns#"
 xmlns="http://purl.org/rss/1.0/"
 xmlns:enc="http://purl.oclc.org/net/rss_2.0/enc#"
 ><foo><title>boogie</title></foo></rdf:RDF>'''

root = ET.fromstring(html)
ns = '{http://purl.org/rss/1.0/}'

titles = root.findall(".//%stitle" % ns)
print titles[0].text

This is working version

answered Jul 7, 2017 at 23:12

gaback

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1 Comment

gaback Over a year ago

you can change this line: titles = root.findall("default:.//title", ns) to titles = root.findall(".//default:title", ns) then it will work. You put default in wrong place

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