如何从其构造函数中设置ElementTree Element的文本字段？或者,在下面的代码中,为什么root.text的第二次打印无？

import xml.etree.ElementTree as ET

root = ET.fromstring("<period units='months'>6</period>")
ET.dump(root)
print root.text

root=ET.Element('period', {'units': 'months'}, text='6')
ET.dump(root)
print root.text

root=ET.Element('period', {'units': 'months'})
root.text = '6'
ET.dump(root)
print root.text

这里输出:

<period units="months">6</period>
6
<period text="6" units="months" />
None
<period units="months">6</period>
6

我是xml解析和Python的新手,所以请耐心等待.我正在使用lxml来解析wiki转储,但我只想要每个页面,它的标题和文本.

现在我有了这个:

from xml.etree import ElementTree as etree

def parser(file_name):
    document = etree.parse(file_name)
    titles = document.findall('.//title')
    print titles

目前,冠军没有返回任何东西.我已经看过像这样的前面的答案:ElementTree findall()返回空列表和lxml文档,但大多数事情似乎都是为解析HTML而定制的.

这是我的XML的一部分:

<mediawiki xmlns="http://www.mediawiki.org/xml/export-0.7/"     xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://www.mediawiki.org/xml/export-0.7/ http://www.mediawiki.org/xml/export-0.7.xsd" version="0.7" xml:lang="en">
  <siteinfo>
  <sitename>Wikipedia</sitename>
<base>http://en.wikipedia.org/wiki/Main_Page</base>
<generator>MediaWiki 1.20wmf9</generator>
<case>first-letter</case>
<namespaces>
  <namespace key="-2" case="first-letter">Media</namespace>
  <namespace key="-1" case="first-letter">Special</namespace>
  <namespace key="0" case="first-letter" />
  <namespace key="1" case="first-letter">Talk</namespace>
  <namespace key="2" case="first-letter">User</namespace>
  <namespace key="3" case="first-letter">User talk</namespace>
  <namespace key="4" case="first-letter">Wikipedia</namespace>
  <namespace key="5" case="first-letter">Wikipedia talk</namespace>
  <namespace key="6" case="first-letter">File</namespace>
  <namespace key="7" case="first-letter">File talk</namespace>
  <namespace key="8" case="first-letter">MediaWiki</namespace>
  <namespace key="9" case="first-letter">MediaWiki talk</namespace>
  <namespace key="10" …

我需要帮助才能理解为什么用xml.etree.ElementTree解析我的xml文件*会产生以下错误.

*我的测试xml文件包含阿拉伯字符.

任务: 打开并解析utf8_file.xml文件.

我的第一次尝试:

import xml.etree.ElementTree as etree
with codecs.open('utf8_file.xml', 'r', encoding='utf-8') as utf8_file:
    xml_tree = etree.parse(utf8_file)

结果1:

UnicodeEncodeError: 'ascii' codec can't encode characters in position 236-238: ordinal not in range(128)

我的第二次尝试:

import xml.etree.ElementTree as etree
with codecs.open('utf8_file.xml', 'r', encoding='utf-8') as utf8_file:
    xml_string = etree.tostring(utf8_file, encoding='utf-8', method='xml')
    xml_tree  = etree.fromstring(xml_string)

结果2:

AttributeError: 'file' object has no attribute 'getiterator'

请解释上述错误并评论可能的解决方案.

在使用ElementTree的Python 2.6中,在特定元素中获取XML(作为字符串)的好方法是什么,比如你在HTML和javascript中可以做什么innerHTML？

这是我开始使用的XML节点的简化示例:

<label attr="foo" attr2="bar">This is some text <a href="foo.htm">and a link</a> in embedded HTML</label>

我想最终得到这个字符串:

This is some text <a href="foo.htm">and a link</a> in embedded HTML

我已经尝试迭代父节点并连接子节点tostring(),但这只给了我子节点:

# returns only subnodes (e.g. <a href="foo.htm">and a link</a>)
''.join([et.tostring(sub, encoding="utf-8") for sub in node])

我可以使用正则表达式破解解决方案,但是希望有一些不那么讨厌的东西:

re.sub("</\w+?>\s*?$", "", re.sub("^\s*?<\w*?>", "", et.tostring(node, encoding="utf-8")))

使用lxml的ElementTree API实现从XML文档中完全删除给定元素很容易,但是我看不到用一些文本一致地替换元素的简单方法.例如,给出以下输入:

input = '''<everything>
<m>Some text before <r/></m>
<m><r/> and some text after.</m>
<m><r/></m>
<m>Text before <r/> and after</m>
<m><b/> Text after a sibling <r/> Text before a sibling<b/></m>
</everything>
'''

...你可以轻松删除每个<r>元素:

from lxml import etree
f = etree.fromstring(data)
for r in f.xpath('//r'):
    r.getparent().remove(r)
print etree.tostring(f, pretty_print=True)

但是,你将如何用文本替换每个元素,以获得输出:

<everything>
<m>Some text before DELETED</m>
<m>DELETED and some text after.</m>
<m>DELETED</m>
<m>Text before DELETED and after</m>
<m><b/>Text after a sibling DELETED Text before a sibling<b/></m>
</everything>

在我看来,这是因为通过与文字ElementTree的API交易.text和 …

给定如下的XML:

<root>
    <element>A</element>
    <element>B</element>
</root>

如何使用ElementTree将元素与内容A匹配并支持XPath？谢谢

我正在使用ElementTree在python中解析XML

import xml.etree.ElementTree as ET 
tree = ET.parse('try.xml')
root = tree.getroot()

我想解析给定目录中的所有'xml'文件.用户应该只输入目录名称,我应该能够遍历目录中的所有文件并逐个解析它们.有人可以告诉我这个方法.我正在使用Linux.

在操作XML时,我希望尽可能忠实地保留注释.

我设法保留了评论,但内容正在进行XML转义.

#!/usr/bin/env python
# add_host_to_tomcat.py

import xml.etree.ElementTree as ET
from CommentedTreeBuilder import CommentedTreeBuilder
parser = CommentedTreeBuilder()

if __name__ == '__main__':
    filename = "/opt/lucee/tomcat/conf/server.xml"

    # this is the important part: use the comment-preserving parser
    tree = ET.parse(filename, parser)

    # get the node to add a child to
    engine_node = tree.find("./Service/Engine")

    # add a node: Engine.Host
    host_node = ET.SubElement(
        engine_node,
        "Host",
        name="local.mysite.com",
        appBase="webapps"
    )
    # add a child to new node: Engine.Host.Context
    ET.SubElement(
        host_node,
        'Context',
        path="",
        docBase="/path/to/doc/base"
    )

    tree.write('out.xml')

#!/usr/bin/env python …

以下测试读取文件,并使用lxml.html为页面生成DOM/Graph的叶节点.

但是,我也试图弄清楚如何从"字符串"获取输入.运用

 lxml.html.fromstring(s)

不起作用,因为这会生成"元素"而不是"ElementTree".

所以,我想弄清楚如何将元素转换为ElementTree.

思考

测试代码::

import lxml.html
from lxml import etree    # trying this to see if needed 
                          # to convert from element to elementtree


  #cmd='cat osu_test.txt'
  cmd='cat o2.txt'
  proc=subprocess.Popen(cmd, shell=True,stdout=subprocess.PIPE)
  s=proc.communicate()[0].strip()

  # s contains HTML not XML text
  #doc = lxml.html.parse(s)
  doc = lxml.html.parse('osu_test.txt')
  doc1 = lxml.html.fromstring(s)

  for node in doc.iter():
  if len(node) == 0:
     print "aaa ",node.tag, doc.getpath(node)
     #print "aaa ",node.tag

  nt = etree.ElementTree(doc1)        <<<<< doesn't work.. so what will??
  for node in nt.iter():
  if len(node) …

我一直在将一些原始xml.etree.ElementTree(ET)代码转换为lxml.etree(lxmlET).幸运的是,两者之间有很多相似之处.但是,我偶然发现了一些我在任何文档中都找不到的奇怪行为.它考虑后代节点的内部表示.

在ET中,iter()用于迭代Element的所有后代,可选地按标记名称进行过滤.因为我在文档中找不到关于此的任何细节,所以我期望lxmlET的类似行为.问题是,从测试我得出结论,在lxmlET中,有一个不同的树内部表示.

在下面的示例中,我迭代树中的节点并打印每个节点的子节点,但此外我还创建了这些子节点的所有不同组合并打印它们.这意味着,如果元素有子元素,('A', 'B', 'C')我会创建更改,即树[('A'), ('A', 'B'), ('A', 'C'), ('B'), ('B', 'C'), ('C')].

# import lxml.etree as ET
import xml.etree.ElementTree as ET
from itertools import combinations
from copy import deepcopy


def get_combination_trees(tree):
    children = list(tree)
    for i in range(1, len(children)):
        for combination in combinations(children, i):
            new_combo_tree = ET.Element(tree.tag, tree.attrib)
            for recombined_child in combination:
                new_combo_tree.append(recombined_child)
                # when using lxml a deepcopy is required to make …