如何在lxml中的find/findall中使用xml名称空间? [英] How do I use xml namespaces with find/findall in lxml?

问题描述

我正在尝试解析OpenOffice ODS电子表格中的内容. ods格式本质上只是一个带有许多文档的zip文件.电子表格的内容存储在"content.xml"中.

I'm trying to parse content in an OpenOffice ODS spreadsheet. The ods format is essentially just a zipfile with a number of documents. The content of the spreadsheet is stored in 'content.xml'.

import zipfile
from lxml import etree

zf = zipfile.ZipFile('spreadsheet.ods')
root = etree.parse(zf.open('content.xml'))

电子表格的内容位于一个单元格中:

The content of the spreadsheet is in a cell:

table = root.find('.//{urn:oasis:names:tc:opendocument:xmlns:table:1.0}table')

我们也可以直接进入这些行:

We can also go straight for the rows:

rows = root.findall('.//{urn:oasis:names:tc:opendocument:xmlns:table:1.0}table-row')

各个元素都知道名称空间:

The individual elements know about the namespaces:

>>> table.nsmap['table']
'urn:oasis:names:tc:opendocument:xmlns:table:1.0'

如何直接在find/findall中使用命名空间?

How do I use the namespaces directly in find/findall?

明显的解决方案不起作用.

The obvious solution does not work.

尝试从表中获取行:

>>> root.findall('.//table:table')
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "lxml.etree.pyx", line 1792, in lxml.etree._ElementTree.findall (src/lxml/lxml.etree.c:41770)
  File "lxml.etree.pyx", line 1297, in lxml.etree._Element.findall (src/lxml/lxml.etree.c:37027)
  File "/usr/lib/python2.6/dist-packages/lxml/_elementpath.py", line 225, in findall
    return list(iterfind(elem, path))
  File "/usr/lib/python2.6/dist-packages/lxml/_elementpath.py", line 200, in iterfind
    selector = _build_path_iterator(path)
  File "/usr/lib/python2.6/dist-packages/lxml/_elementpath.py", line 184, in _build_path_iterator
    selector.append(ops[token[0]](_next, token))
KeyError: ':'

推荐答案

如果root.nsmap包含table名称空间前缀，则可以:

If root.nsmap contains the table namespace prefix then you could:

root.xpath('.//table:table', namespaces=root.nsmap)

findall(path)接受{namespace}name语法而不是namespace:name.因此，在将path传递给findall()之前，应使用名称空间字典将其预处理为{namespace}name形式.

findall(path) accepts {namespace}name syntax instead of namespace:name. Therefore path should be preprocessed using namespace dictionary to the {namespace}name form before passing it to findall().

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