如何将lxml xpath函数导入默认名称空间? [英] How to import lxml xpath functions to default namespace?

问题描述

这是 lxml文档中的示例:

>>> regexpNS = "http://exslt.org/regular-expressions"
>>> find = etree.XPath("//*[re:test(., '^abc$', 'i')]",
...                    namespaces={'re':regexpNS})

>>> root = etree.XML("<root><a>aB</a><b>aBc</b></root>")
>>> print(find(root)[0].text)
aBc

我想将re:test()函数导入默认名称空间，以便我可以在没有前缀re:的情况下调用它.我该怎么做?谢谢！

I want to import re:test() function to default namespace, so that I can call it without prefix re:. How can I do it? Thanks!

推荐答案

您可以将函数放在空函数名称空间中:

You can put a function in the empty function namespace:

functionNS = etree.FunctionNamespace(None)
functionNS['test'] = lambda context, nodes, *args: print(context, nodes, args)

这样做，新的test函数已经用空的名称空间前缀注册，这意味着您可以像这样使用它:

By doing so, the new test function is already registered with the empty namespace prefix, that means you can use it like this:

root.xpath("//*[test(., 'arg1', 'arg2')]")

不幸的是，被称为的功能对于"{http://exslt.org/regular-expressions}test"，它不能从python中获得，只能从用C实现的lxml扩展中获得，因此您不能简单地将其分配给functionNS['test'].

Unfortunately the function that is called for "{http://exslt.org/regular-expressions}test" isn't available from python, only from within the lxml extension implemented in C, so you can't simply assign it to functionNS['test'].

这意味着您需要在python中重新实现它，以将其分配给空函数名称空间...

That means you'd need to reimplement it in python to assign it to the empty function namespace...

如果不必为您省去键入三个字符的麻烦，则可以使用此技巧使名称空间的re前缀变为全局:

If that's not worth the trouble for you to spare you typing three characters, you could use this trick to make the re prefix for the namespace global:

etree.FunctionNamespace("http://exslt.org/regular-expressions").prefix = 're'

那么至少您不需要为每个xpath表达式传递名称空间dict.

Then at least you don't need to pass the namespaces dict for each xpath expression.

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