The problem is that findall only searches the immediate descendants of an element if it is given a tag name. You need to give it an XPath expression that will find track anywhere in the tree beneath it. So the following should work, for example: api_results = ElementTree.parse(response).findall(‘.//track’)
This compare function works for me: def elements_equal(e1, e2): if e1.tag != e2.tag: return False if e1.text != e2.text: return False if e1.tail != e2.tail: return False if e1.attrib != e2.attrib: return False if len(e1) != len(e2): return False return all(elements_equal(c1, c2) for c1, c2 in zip(e1, e2))
Although the docs say otherwise, I only was able to get an <?xml> declaration by specifying both the xml_declaration and the encoding. You have to declare nodes in the namespace you’ve registered to get the namespace on the nodes in the file. Here’s a fixed version of your code: from xml.etree import ElementTree as ET … Read more
You can use the function toprettyxml() from xml.dom.minidom in order to do that: def prettify(elem): “””Return a pretty-printed XML string for the Element. “”” rough_string = ElementTree.tostring(elem, ‘utf-8’) reparsed = minidom.parseString(rough_string) return reparsed.toprettyxml(indent=”\t”) The idea is to print your Element in a string, parse it using minidom and convert it again in XML using the … Read more
As far as I know the solution that better suits your needs is to write a pure Python custom rendering using the features exposed by xml.etree.ElementTree. Here is one possible solution: from xml.etree import ElementTree as ET from re import findall, sub def render(root, buffer=””, namespaces=None, level=0, indent_size=2, encoding=’utf-8′): buffer += f'<?xml version=”1.0″ encoding=”{encoding}” ?>\n’ … Read more
The code for creating a dictionary with all the declared namespaces can be made quite simple. This is all that is needed: import xml.etree.ElementTree as ET my_namespaces = dict([node for _, node in ET.iterparse(‘file.xml’, events=[‘start-ns’])]) You don’t need to use StringIO or open(). Just provide the XML filename as an argument to iterparse(). Each item … Read more
If you have third party module that wants to use ElementTree (and XMLTreeBuilder by dependency) you can change ElementTree’s XMLTreeBuilder definition to the one provided by SimpleXMLTreeBuilder like so: from xml.etree import ElementTree # part of python distribution from elementtree import SimpleXMLTreeBuilder # part of your codebase ElementTree.XMLTreeBuilder = SimpleXMLTreeBuilder.TreeBuilder Now ElementTree will always use … Read more
What the code you posted is doing is combining all the elements regardless of whether or not an element with the same tag already exists. So you need to iterate over the elements and manually check and combine them the way you see fit, because it is not a standard way of handling XML files. … Read more
AFAIK ElementTree does not support XPath. Has it changed? Anyway, you can use lxml and the following XPath expression: import lxml.etree doc = lxml.etree.parse(‘t.xml’) print doc.xpath(‘//element[text()=”A”]’)[0].text print doc.xpath(‘//element[text()=”A”]’)[0].tag The result will be: A element
You can pass the value returned by urlopen() directly to GzipFile() and in turn you can pass it to ElementTree methods such as iterparse(): #!/usr/bin/env python3 import xml.etree.ElementTree as etree from gzip import GzipFile from urllib.request import urlopen, Request with urlopen(Request(“http://smarkets.s3.amazonaws.com/oddsfeed.xml”, headers={“Accept-Encoding”: “gzip”})) as response, \ GzipFile(fileobj=response) as xml_file: for elem in getelements(xml_file, ‘interesting_tag’): process(elem) … Read more