使用xpath查询将xpath正则表达式捕获组作为字符串返回

Question

CONTEXT

Supposing the following HTML

....
<p>Whatever</p>
<div>Whatever DIV78232 Everwhat</div>
....

Question:

How could I return a plain text string containing DIVnnnnn, where nnnnn represents any digits.

My investigation so far:

The xPath replace() function will replace a pattern found inside the current DOM.

replace(.,'.*?(DIV\d+).*','$1') => DIV78232

Why am I blocked?

Because I want the query to return the "DIV78232" as a string, without actually replacing it in the DOM at all, just as it would return "Whatever" for the query /p/text() [I am trying all this on the FirePath firefox-extension]

Note: According to the official DOCS

"replace() Returns the value of the first argument with every substring matched by the regular expression that is the value of the second argument replaced by the replacement string that is the value of the third argument."

FINAL PURPOSE:

My final purpose is to get the (string) IMAGE URL that matches '.*?image:.*?"(.+?)".*' from this (which is inside the HTML):

In this case, the query //*[matches(.,'.*?image:.*?"(.+?)".*','i')] returns the whole node, but I only want the first Capturing Group, which would be the IMAGE URL.

<script>...vp&output=xml_vast2&unviewed_position_start=1&
url='+encodeURIComponent(location.href)+'
description_url='+encodeURIComponent(location.href)+'&
image:   "https://domain.com/xxxxxxx/public_images/2015.12/article/56797be1c46188ac438b45c3.jpg", // stretching: 'fi..</script>

Answer 1

Took me a long while, but this is the result I got by combinating replace() and tokenize()

tokenize(replace(.,'.*?image:.*?"(.+?)".*?',':@:$1:@:'),':@:')[2]

Returns the image URL in the snippet above mentioned.

Why/How does this work?

• Replace() matches the image and wraps the capturing group with my own token separator ':@:' (Could be anything original)
• Tokenize() splits the replaced string in 3 parts, being the second one the capturing group I was looking for. (It will be three parts because it is highly improbable that the document will contain ':@:' anywhere else)

Is there any faster way to achieve this?

Thanks. All the best. Peace.