get XPATH for all the nodes

2019-02-03 00:17发布

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is there any library that can give me XPATH for all the nodes in HTML page ?

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太酷不给撩
2楼-- · 2019-02-03 00:47

In case this is helpful for someone else, if you're using python/lxml, you'll first need to have a tree, and then query that tree with the XPATH paths that Dimitre lists above.

To get the tree:

import lxml
from lxml import html, etree

your_webpage_string = "<html><head><title>test<body><h1>page title</h3>"
bad_html = lxml.html.fromstring(your_webpage_string)
good_html = etree.tostring(root, pretty_print=True).strip()
your_tree = etree.fromstring(good_html)
all_xpaths = your_tree.xpath('//*')

On the last line, replace '//*' with whatever xpath you want. all_xpaths is now a list which looks like this:

[<Element html at 0x7ff740b24b90>,
 <Element head at 0x7ff740b24d88>,
 <Element title at 0x7ff740b24dd0>,
 <Element body at 0x7ff740b24e18>,
 <Element h1 at 0x7ff740b24e60>]
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别忘想泡老子
3楼-- · 2019-02-03 00:51

is there any library that can give me XPATH for all the nodes in HTML page

Yes, if this HTML page is a well-formed XML document.

Depending on what you understand by "node"...

//*

selects all the elements in the document.

/descendant-or-self::node()

selects all elements, text nodes, processing instructions, comment nodes, and the root node /.

//text()

selects all text nodes in the document.

//comment()

selects all comment nodes in the document.

//processing-instruction()

selects all processing instructions in the document.

//@*

selects all attribute nodes in the document.

//namespace::*

selects all namespace nodes in the document.

Finally, you can combine any of the above expressions using the union (|) operator.

Thus, I believe that the following expression really selects "all the nodes" of any XML document:

/descendant-or-self::node() | //@* | //namespace::*
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