Reputation: 1091
I have a an xml document like the following:
<?xml version="1.0" encoding="UTF-8" standalone="no"?>
<project xmlns="http://maven.apache.org/POM/4.0.0" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://maven.apache.org/POM/4.0.0 http://maven.apache.org/xsd/maven-4.0.0.xsd">
<modelVersion>4.0.0</modelVersion>
<parent>
<groupId>company</groupId>
<artifactId>art-id</artifactId>
<version>RELEASE</version>
</parent>
<properties>
<tomcat.username>admin</tomcat.username>
<tomcat.password>admin</tomcat.password>
</properties>
<dependencies>
<dependency>
<groupId>asdf</groupId>
<artifactId>asdf</artifactId>
<version>[3.8,)</version>
</dependency>
<dependency>
<groupId>asdf</groupId>
<artifactId>asdf</artifactId>
<version>[4.1,)</version>
</dependency>
</dependencies>
how can I delete the entire node "dependencies"?
I have looked at other questions and answers on stackoverflow and what is different about is the namespace aspect of this xml, and the other questions ask to delete a subelement like "dependency" while I want to delete the whole node "dependencies." Is there an easy way using lxml to delete the entire node?
The following gives a 'NoneType' object has no attribute 'remove' error:
from lxml import etree as ET
tree = ET.parse('pom.xml')
namespace = '{http://maven.apache.org/POM/4.0.0}'
root = ET.Element(namespace+'project')
root.find(namespace+'dependencies').remove()
Upvotes: 1
Views: 2299
Reputation: 184
First, grab the root node. Since it is <project ... >
(vs <project .../>
) the "parent" element of dependencies
is project
. Example from the documentation:
import xml.etree.ElementTree as ET
tree = ET.parse('country_data.xml')
root = tree.getroot()
Once you have the root, check root.tag()
, it should be "project".
Then do root.remove(root.find('dependencies'))
, where root
is the project
node.
If it were <project .../>
then it would be invalid XML since there must be a root element. I can see exactly where you are coming from, though.
Upvotes: 1
Reputation: 180411
You can create a dict mapping for your namespace(s), find the node then call root.remove passing the node, you don't call .remove on the node:
x = """<?xml version="1.0" encoding="UTF-8" standalone="no"?>
<project xmlns="http://maven.apache.org/POM/4.0.0" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://maven.apache.org/POM/4.0.0 http://maven.apache.org/xsd/maven-4.0.0.xsd">
<modelVersion>4.0.0</modelVersion>
<parent>
<groupId>company</groupId>
<artifactId>art-id</artifactId>
<version>RELEASE</version>
</parent>
<properties>
<tomcat.username>admin</tomcat.username>
<tomcat.password>admin</tomcat.password>
</properties>
<dependencies>
<dependency>
<groupId>asdf</groupId>
<artifactId>asdf</artifactId>
<version>[3.8,)</version>
</dependency>
<dependency>
<groupId>asdf</groupId>
<artifactId>asdf</artifactId>
<version>[4.1,)</version>
</dependency>
</dependencies>
</project>"""
import lxml.etree as et
from StringIO import StringIO
tree = et.parse(StringIO(x))
root =tree.getroot()
nsmap = {"mav":"http://maven.apache.org/POM/4.0.0"}
root.remove(root.find("mav:dependencies", namespaces=nsmap))
print(et.tostring(tree))
Which would give you:
<project xmlns="http://maven.apache.org/POM/4.0.0" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://maven.apache.org/POM/4.0.0 http://maven.apache.org/xsd/maven-4.0.0.xsd">
<modelVersion>4.0.0</modelVersion>
<parent>
<groupId>company</groupId>
<artifactId>art-id</artifactId>
<version>RELEASE</version>
</parent>
<properties>
<tomcat.username>admin</tomcat.username>
<tomcat.password>admin</tomcat.password>
</properties>
</project>
Upvotes: 1