问题
I have an XML document in which I want to search for some elements and if they match some criteria I would like to delete them
However, I cannot seem to be able to access the parent of the element so that I can delete it
file = open('test.xml', "r")
elem = ElementTree.parse(file)
namespace = "{http://somens}"
props = elem.findall('.//{0}prop'.format(namespace))
for prop in props:
type = prop.attrib.get('type', None)
if type == 'json':
value = json.loads(prop.attrib['value'])
if value['name'] == 'Page1.Button1':
#here I need to access the parent of prop
# in order to delete the prop
Is there a way I can do this?
Thanks
回答1:
You can remove child elements with the according remove method. To remove an element you have to call its parents remove method. Unfortunately Element does not provide a reference to its parents, so it is up to you to keep track of parent/child relations (which speaks against your use of elem.findall())
A proposed solution could look like this:
root = elem.getroot()
for child in root:
if child.name != "prop":
continue
if True:# TODO: do your check here!
root.remove(child)
PS: don't use prop.attrib.get(), use prop.get(), as explained here.
回答2:
You could use xpath to select an Element's parent.
file = open('test.xml', "r")
elem = ElementTree.parse(file)
namespace = "{http://somens}"
props = elem.findall('.//{0}prop'.format(namespace))
for prop in props:
type = prop.get('type', None)
if type == 'json':
value = json.loads(prop.attrib['value'])
if value['name'] == 'Page1.Button1':
# Get parent and remove this prop
parent = prop.find("..")
parent.remove(prop)
http://docs.python.org/2/library/xml.etree.elementtree.html#supported-xpath-syntax
Except if you try that it doesn't work: http://elmpowered.skawaii.net/?p=74
So instead you have to:
file = open('test.xml', "r")
elem = ElementTree.parse(file)
namespace = "{http://somens}"
search = './/{0}prop'.format(namespace)
# Use xpath to get all parents of props
prop_parents = elem.findall(search + '/..')
for parent in prop_parents:
# Still have to find and iterate through child props
for prop in parent.findall(search):
type = prop.get('type', None)
if type == 'json':
value = json.loads(prop.attrib['value'])
if value['name'] == 'Page1.Button1':
parent.remove(prop)
It is two searches and a nested loop. The inner search is only on Elements known to contain props as first children, but that may not mean much depending on your schema.
回答3:
Using the fact that every child must have a parent, I'm going to simplify @kitsu.eb's example. f using the findall command to get the children and parents, their indices will be equivalent.
file = open('test.xml', "r")
elem = ElementTree.parse(file)
namespace = "{http://somens}"
search = './/{0}prop'.format(namespace)
# Use xpath to get all parents of props
prop_parents = elem.findall(search + '/..')
props = elem.findall('.//{0}prop'.format(namespace))
for prop in props:
type = prop.attrib.get('type', None)
if type == 'json':
value = json.loads(prop.attrib['value'])
if value['name'] == 'Page1.Button1':
#use the index of the current child to find
#its parent and remove the child
prop_parents[props.index[prop]].remove(prop)
回答4:
I know this is an old thread but this kept popping up while I was trying to figure out a similar task. I did not like the accepted answer for two reasons:
1) It doesn't handle multiple nested levels of tags.
2) It will break if multiple xml tags are deleted in the same level one-after-another. Since each element is an index of Element._children you shouldn't delete while forward iterating.
I think a better more versatile solution is this:
import xml.etree.ElementTree as et
file = 'test.xml'
tree = et.parse(file)
root = tree.getroot()
def iterator(parents, nested=False):
for child in reversed(parents):
if nested:
if len(child) >= 1:
iterator(child)
if True: # Add your entire condition here
parents.remove(child)
iterator(root, nested=True)
For the OP, this should work - but I don't have the data you're working with to test if it's perfect.
import xml.etree.ElementTree as et
file = 'test.xml'
tree = et.parse(file)
namespace = "{http://somens}"
props = tree.findall('.//{0}prop'.format(namespace))
def iterator(parents, nested=False):
for child in reversed(parents):
if nested:
if len(child) >= 1:
iterator(child)
if prop.attrib.get('type') == 'json':
value = json.loads(prop.attrib['value'])
if value['name'] == 'Page1.Button1':
parents.remove(child)
iterator(props, nested=True)
回答5:
A solution using lxml module
from lxml import etree
root = ET.fromstring(xml_str)
for e in root.findall('.//{http://some.name.space}node'):
parent = e.getparent()
for child in parent.find('./{http://some.name.space}node'):
try:
parent.remove(child)
except ValueError:
pass
回答6:
I like to use an XPath expression for this kind of filtering. Unless I know otherwise, such an expression must be applied at the root level, which means I can't just get a parent and apply the same expression on that parent. However, it seems to me that there is a nice and flexible solution that should work with any supported XPath, as long as none of the sought nodes is the root. It goes something like this:
root = elem.getroot()
# Find all nodes matching the filter string (flt)
nodes = root.findall(flt)
while len(nodes):
# As long as there are nodes, there should be parents
# Get the first of all parents to the found nodes
parent = root.findall(flt+'/..')[0]
# Use this parent to remove the first node
parent.remove(nodes[0])
# Find all remaining nodes
nodes = root.findall(flt)
来源：https://stackoverflow.com/questions/6847263/search-and-remove-element-with-elementtree-in-python