我有这个文档,我需要解析它并得到一个XML相当。基本上我需要一个ElementTree类型的对象,但它没有发生。我试过许多不同的组合,但我还没有弄明白。 我是这样做的:
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,所以我使用下面的方法来获取一个元素。在
还有这个:
tree = ET.ElementTree(doc_xml) #type(tree) is 'ElementTree'
type(tree)
我以为这很管用,但当我这么做的时候:
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
属于ElementTree类型。我如何解决这个问题?在
有了这条线
通过解析以字符串形式给出的XML文档,可以得到一个
^{pr2}$Element
实例。您可以迭代此实例:当你这么做的时候
通过提供一个字符串作为参数,可以创建一个
ElementTree
实例。这不会导致错误消息,但尝试遍历树失败,因为它不是“真正的”树(在本例中,不会解析XML)。在如果您需要一个
ElementTree
实例,我建议您这样做:相关问题 更多 >
编程相关推荐