Question

langs对象是通过解析此文件生成的lxml对象：http://www.loc.gov/standards/codelists/languages.xml

此xpath有效：

langs.node.xpath("//lang:language[lang:name='English']", namespaces={'lang':'info:lc/xmlns/codelist-v1'})[0].findtext('lang:name', namespaces={'lang': 'info:lc/xmlns/codelist-v1'})

当我添加 |的附加条件时lang：code =＆＃39;英语＆＃39; 喜欢这样：

langs.node.xpath("//lang:language[lang:name='English' | lang:code='English']", namespaces={'lang':'info:lc/xmlns/codelist-v1'})[0].findtext('lang:name', namespaces={'lang': 'info:lc/xmlns/codelist-v1'})

我收到错误：

Traceback (most recent call last):
  File "<console>", line 1, in <module>
  File "lxml.etree.pyx", line 1509, in lxml.etree._Element.xpath (src/lxml /lxml.etree.c:50717)
  File "xpath.pxi", line 318, in lxml.etree.XPathElementEvaluator.__call__ (src/lxml/lxml.etree.c:145969)
  File "xpath.pxi", line 238, in lxml.etree._XPathEvaluatorBase._handle_result (src/lxml/lxml.etree.c:144977)
  File "xpath.pxi", line 223, in lxml.etree._XPathEvaluatorBase._raise_eval_error  (src/lxml/lxml.etree.c:144785)
XPathEvalError: Invalid type

Answer 1

正确的语法是or，而不是|：

nsmap={'lang': 'info:lc/xmlns/codelist-v1'}
langs.node.xpath("//lang:language[lang:name='English' or lang:code='English']",
  namespaces=nsmap)[0].findtext('lang:name', namespaces=nsmap)

Answer 2

再过10分钟的谷歌搜索，我得到了答案：

我将 | 更改为或

langs.node.xpath("//lang:language[lang:name='English' or lang:code='English']", namespaces={'lang':'info:lc/xmlns/codelist-v1'})[0].findtext('lang:name', namespaces={'lang': 'info:lc/xmlns/codelist-v1'})

OR条件不起作用的lxml xpath过滤器

2 个答案: