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@IanHopkinson
Created November 24, 2015 19:42
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Examples of xpath queries using lxml in python
#!/usr/bin/env python
# encoding: utf-8
import lxml.etree
import lxml.html
import requests
xml_sample = """<?xml version="1.0" encoding="UTF-8"?>
<foo:Results xmlns:foo="http://www.foo.com" xmlns="http://www.bah.com">
<foo:Recordset setCount="2">
<foo:Record setEntry="0">
<foo:Title>First title</foo:Title>
</foo:Record>
<foo:Record setEntry="1">
<foo:Title>Second title</foo:Title>
</foo:Record>
<Record setEntry="2">
<Title>Third title</Title>
</Record>
<Record setEntry="3">
<Title>Fourth title</Title>
</Record>
</foo:Recordset>
</foo:Results>
""".encode("utf-8")
def main():
print("Demonstrating xpath on HTML")
print("===========================")
r = requests.get("http://www.ianhopkinson.org.uk")
root = lxml.html.fromstring(r.content)
title = root.xpath('/html/body/div/div/div[2]/h1')
print("My blog title is: '{}'".format(title[0].text.strip()))
title = root.xpath('//div[2]/h1')
print("We can use the // shortcut to get the same thing more easily: '{}'".format(title[0].text_content().strip()))
ids = root.xpath('//li/@id')
print("We can get the id attributes of all the <li> elements. There are {} of them, the first one is {}".format(len(ids), ids[0]))
tagcloud = root.xpath('//*[@class="tagcloud"]')
print("We can get the parent element of the tagcloud using an attribute selector: {}".format(tagcloud))
title = root.xpath("//h1[contains(., 'SomeBeans')]")
print("Another way to get the title is to select by element text content: '{}'".format(title[0].text.strip()))
subtitle = root.xpath('//h1[contains(@class,"header_title")]/../h2')
print("We can use the .. operator is select the subtitle: '{}'".format(subtitle[0].text.strip()))
subtitle = root.xpath('//h1[contains(@class,"header_title")]/following-sibling::h2')
print("Or we can use following-sibling to same effect: '{}'".format(subtitle[0].text.strip()))
print("\nDemonstrating xpath on XML")
print("============================")
print("Processing XML is pretty similar except for namespaces")
namespace = "http://www.foo.com"
namespace_c = "{" + namespace + "}"
NSMAP = {"foo": namespace}
root = lxml.etree.fromstring(xml_sample)
record_count = root.xpath('//@setCount')[0]
print("Attributes are easy, this is the @setCount: {}".format(record_count))
print("These are the elements defined by the XML string at the top of this program:")
for i, element in enumerate(root.getiterator()):
print(element.tag)
print("We can select elements by defining a namespace in our queries")
records = root.xpath('//foo:Title', namespaces = {"foo": "http://www.foo.com"})
for record in records:
print(record.text)
print("Without defining the default namespace, we get nothing")
records = root.xpath('//Title')
for record in records:
print(record.text)
print("With the default namespace, we get something")
records = root.xpath('//bah:Title', namespaces = {"bah": "http://www.bah.com"})
for record in records:
print("Element name: {}, element text '{}'".format(record.tag, record.text))
if __name__ == "__main__":
main()
@IanHopkinson
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The key point is that we can have an xml file that does not have prefixes for elements, they implicitly take the default namespace. When we query the xml we have to specify the default namespace. As I recall my mistake in the past was believing that my query would pick up the default namespace from the XML file it was querying, and that is not the case.

@lsloan
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lsloan commented Oct 26, 2021

What is the purpose of the namespace variables on lines 59–61? They're not used anywhere else that I can see.

@IanHopkinson
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@lsloan I think that is probably a hangover from an earlier version of the code, in this version it serves no purpose. The lxml documentation uses that style of namespace definition, I probably intended to use it down at line 74 and then forgot!

@lsloan
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lsloan commented Nov 1, 2021

Gotcha. I thought that might be the case.

I forked your gist and made some changes. Then I added on some other examples of processing XML that contains QTI (Question & Test Interoperability) data. Experimenting with lxml.etree, I found that the default, unnamed namespace in the XML is available in the tree's data in nsmap[None]. See my lxml-test-etree.py, line 11

    defaultNamespace = {'_': root.nsmap[None]}

I found that naming it _ makes it convenient to refer to it in the XPath statement, as on line 23

    items = root.xpath('//_:item', namespaces=defaultNamespace)

I wanted to get that namespace used by default when xpath() is called. I tried setting the key to None or using root.nsmap itself, but those caused an error ("TypeError: empty namespace prefix is not supported in XPath").

I'd like to not need to use the _: prefix for the element name, but at least it's minimally obtrusive. Trying to set a truly default namespace is a lost cause, apparently. As written in the lxml FAQ, "How can I specify a default namespace for XPath expressions?". The short answer: "You can't." 🤷

As it turns out, I may prefer using lxml.objectify rather than lxml.etree, but I need to investigate a little more before I know for sure. See my lxml-test-objectify.py, for example.

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