Get last page number from HTML - bash

Question

I have a page pagination that look like this in HTML:
<span class="nav">Go to <b>1</b>, 2, 3, 4, 5 Next</span>
What I want to get is the last page number (so in this example "5"). How I can do it in Bash? Thank you for help.

Answer 1

As a solution that only considers numbers given as the text associated with links inside of <span class="nav"> (assuming in.html as your input file):
xmllint --html --xmlout - <in.html \
| xmlstarlet sel -t -m '//span[#class="nav"]//a' -v 'text()' -n \
| egrep '^[[:digit:]]+$' \
| sort -n \
| tail -n 1
This uses xmllint (included with modern Linux distributions) to convert your HTML to XML, and XMLStarlet (not always included, but generally packaged for common distributions) to search that XML.

Answer 2

This assumes that the HTML is always conform your input:
sed 's/page-/\n/g' | sort -n | tail -1 | sed 's/.html.*//'
(sed 's/page-/\n/g' puts a newline just before the page number. sort -n sorts numerically; the lines that do not start with a page number get sorted on top. tail -1 selects the line with the highest page number and the sed 's/.html.*//' strips-off all the non-pagenumber stuff)

Answer 3

If there are only numbers of the pages only numbers in the text, then you can do it like following
egrep '[0-9]+' -o | sort -r -n | head -1
it will match numbers from the text, than sort it and take the first one (highest). You can modify the regexp if you want so be more specific. Better approach would be deffinitely possible in python using BeautifulSoup4 where you can traverse through the DOM like in jQuery.
EDIT added -n to the command (+1 #CharlesDuffy)