Question

我有这个文档，我需要解析并获得XML等价物。基本上我需要一个ElementTree类型的对象，但它没有发生。我尝试了很多不同的组合，但我还没弄清楚。这是我做的：

import xml.etree.ElementTree as ET
z = zf.ZipFile("INTRODUCTION.docx")
doc_xml = z.read("word/document.xml")
print doc_xml           #type(doc_xml) is str  

<?xml version="1.0" encoding="UTF-8" standalone="yes"?>
<w:document xmlns:wpc="http://schemas.microsoft.com/office/word/2010/wordprocessingCanvas" xmlns:mc="http://schemas.openxmlformats.org/markup-compatibility/2006" xmlns:o="urn:schemas-microsoft-com:office:office" xmlns:r="http://schemas.openxmlformats.org/officeDocument/2006/relationships" xmlns:m="http://schemas.openxmlformats.org/officeDocument/2006/math" xmlns:v="urn:schemas-microsoft-com:vml" xmlns:wp14="http://schemas.microsoft.com/office/word/2010/wordprocessingDrawing" xmlns:wp="http://schemas.openxmlformats.org/drawingml/2006/wordprocessingDrawing" xmlns:w10="urn:schemas-microsoft-com:office:word" xmlns:w="http://schemas.openxmlformats.org/wordprocessingml/2006/main" xmlns:w14="http://schemas.microsoft.com/office/word/2010/wordml" xmlns:w15="http://schemas.microsoft.com/office/word/2012/wordml" xmlns:wpg="http://schemas.microsoft.com/office/word/2010/wordprocessingGroup" xmlns:wpi="http://schemas.microsoft.com/office/word/2010/wordprocessingInk" xmlns:wne="http://schemas.microsoft.com/office/word/2006/wordml" xmlns:wps="http://schemas.microsoft.com/office/word/2010/wordprocessingShape" mc:Ignorable="w14 w15 wp14"><w:body><w:p w:rsidR="00470EEF" w:rsidRDefault="00456755"><w:pPr><w:rPr><w:b/></w:rPr></w:pPr><w:r w:rsidRPr="00456755"><w:rPr><w:b/></w:rPr><w:t>INTRODUCTION</w:t></w:r></w:p><w:p w:rsidR="00456755" w:rsidRDefault="00456755"><w:r w:rsidRPr="00456755"><w:t>This is a test document for xml</w:t></w:r><w:r><w:t>.</w:t></w:r></w:p><w:p w:rsidR="00456755" w:rsidRDefault="00456755"><w:proofErr w:type="spellStart"/><w:proofErr w:type="gramStart"/><w:r><w:t>Lets</w:t></w:r><w:proofErr w:type="spellEnd"/><w:proofErr w:type="gramEnd"/><w:r><w:t xml:space="preserve"> see how this works.</w:t></w:r></w:p><w:p w:rsidR="00456755" w:rsidRDefault="00456755"/><w:p w:rsidR="00456755" w:rsidRDefault="00456755"/><w:p w:rsidR="00456755" w:rsidRDefault="00456755"><w:pPr><w:rPr><w:b/></w:rPr></w:pPr><w:r w:rsidRPr="00456755"><w:rPr><w:b/></w:rPr><w:t>Conclusion</w:t></w:r></w:p><w:p w:rsidR="00456755" w:rsidRPr="00456755" w:rsidRDefault="00456755"><w:r w:rsidRPr="00456755"><w:t>It should hopefully</w:t></w:r><w:r><w:t>..</w:t></w:r><w:bookmarkStart w:id="0" w:name="_GoBack"/><w:bookmarkEnd w:id="0"/></w:p><w:sectPr w:rsidR="00456755" w:rsidRPr="00456755"><w:pgSz w:w="11906" w:h="16838"/><w:pgMar w:top="1440" w:right="1440" w:bottom="1440" w:left="1440" w:header="708" w:footer="708" w:gutter="0"/><w:cols w:space="708"/><w:docGrid w:linePitch="360"/></w:sectPr></w:body></w:document>

由于doc_xml是string类型，我使用以下内容来获取Element。

rooted = ET.fromstring(doc_xml)    #type(rooted) is 'Element'
type(rooted)

这也是：

tree = ET.ElementTree(doc_xml)  #type(tree) is 'ElementTree'
type(tree)

我认为这样可行，但当我这样做时：

for branch in tree.iter():
    print branch  
---------------------------------------------------------------------------
AttributeError                            Traceback (most recent call last)
<ipython-input-83-d503315fb5e6> in <module>()
----> 1 for branch in tree.iter():
      2     print branch

C:\Anaconda\lib\xml\etree\ElementTree.pyc in iter(self, tag)
    671     def iter(self, tag=None):
    672         # assert self._root is not None
--> 673         return self._root.iter(tag)
    674 
    675     # compatibility

AttributeError: 'str' object has no attribute 'iter'

变量tree是ElementTree类型。我该如何解决这个问题？

Answer 1

有了这一行，

rooted = ET.fromstring(doc_xml)

通过解析以字符串形式给出的XML文档来获取Element实例。您可以遍历此实例：

for branch in rooted.iter():
    print branch

当你这样做时，

tree = ET.ElementTree(doc_xml)

通过提供字符串作为参数来创建ElementTree实例。这不会导致错误消息，但尝试迭代树失败，因为它不是“真正的”树（在这种情况下不解析XML）。

如果您需要ElementTree个实例，我建议您这样做：

import xml.etree.ElementTree as ET
import zipfile as zf

z = zf.ZipFile("INTRODUCTION.docx")
f = z.open("word/document.xml")   # a file-like object
tree = ET.parse(f)                # an ElementTree instance

for elem in tree.iter():
    print elem

使用元素树模块解析docx

1 个答案: