Insert and renumber endnotes

[chouette]

Hello all, I'm new to the forum and have been using Calibre for e-book conversion and editing. My HTML skills are rudimentary, but I can usually manage to polish an auto-generated EPUB.

Is there a way for me insert a totally new endnote (from a revision) and its corresponding back link into an e-book without having to manually renumber all the (hundreds of) endnotes that follow? Thanks!

[kovidgoyal]

There's no builtin tool for it, but you can use the function mode of search and replace to create something, if you are willing to write a small function in python. See https://manual.calibre-ebook.com/function_mode.html

[chouette]

Thank you so much for your quick response! I've never written a function before. Perhaps someone would be willing to give me a quick tutorial using this particular example? Thanks again.

[lomkiri]

If, e.g., your note-anchor is something like this :

<a class="noteAnchor" href="../Text/footnotes.xhtml#footnote-0003">[3]</a>

and you want to insert the note [80] (i.e. the old n. 80 (and over) will be renumbered)

— Make a backup of you book
— Put as a search string (you'll have to adapt it to the form of the note-anchors of your ebook) :

<a[^>]+>\[\K(\d+)(\])

— Set the search as regex-function and create the function with the code:

def replace(match, number, file_name, metadata, dictionaries, data, functions, *args, **kwargs):
    start_at = 80
    if (n := int(match.group(1))) >= start_at:
        return str(n+1) + match.group(2)
    return match.group(0)

Of course, you must change the var. "start_at" to feet your needs.
— Replace all
— Insert your own note as n. 80

Same process for the notes themselves, just adapt the regex to their form..
If you don't know how to adapt the regexes to your case, give an example of the note-anchor and of the body of the note.

[chouette]

Originally Posted by lomkiri

If, e.g., your note-anchor is something like this :

<a class="noteAnchor" href="../Text/footnotes.xhtml#footnote-0003">[3]</a>

and you want to insert the note [80] (i.e. the old n. 80 (and over) will be renumbered)

— Make a backup of you book
— Put as a search string (you'll have to adapt it to the form of the note-anchors of your ebook) :

<a[^>]+>\[\K(\d+)(\])

— Set the search as regex-function and create the function with the code:

def replace(match, number, file_name, metadata, dictionaries, data, functions, *args, **kwargs):
    start_at = 80
    if (n := int(match.group(1))) >= start_at:
        return str(n+1) + match.group(2)
    return match.group(0)

Of course, you must change the var. "start_at" to feet your needs.
— Replace all
— Insert your own note as n. 80

Same process for the notes themselves, just adapt the regex to their form..
If you don't know how to adapt the regexes to your case, give an example of the note-anchor and of the body of the note.

Thank you so much! My note link is as follows:

<a id="back_note_3" href="notes.html#note_3" title="3" class="noteref">3</a>

And the note itself is like this:

<dl id="note_3" class="footnote"><dt class="calibre42">[<a href="ch3.html#back_note_3" title="3" class="calibre43">←3</a>]</dt><dd class="calibre44"><p class="block_91"> Note text goes here.</p></dd></dl>

[lomkiri]

Well, if all your note-anchors and note-bodies are on the same form, those two regexes will do the job:

Note-anchors:
class="noteref">\K(\d+)(</a>)
note bodies:
\[<a href=[^#]+#back_note_\d+" title="\d+" class="calibre\d+">←\K(\d+)(</a>\])

The function is still the same for both form.

Although it's possible (and not difficult) to adapt the numbers inside the tags (back_note_3, note_3), for the sake of simplicity I didn't touch to them, only to the displayed numbers. When you'll add your own note, e.g. 80, put 80a as numbers in the tags, to avoid duplicated links:
<a id="back_note_80a" href="notes.html#note_80a" title="80a" class="noteref">80</a>

Don't forget to:
1) back-up you book first
2) change the value of "start_at" in the function
3) See the differences after a replace all, to check if it went as you were expecting
4) Check the book with the bug-icon or with F7
5) Check some notes liink to see if they are still working properly

[ugosan]

Hello. I tried this function, but I can't set it: when I insert it copy&paste in "create function" it returns me: "The code you created is not valid Python code, with error: invalid sintax (name, line 3)".
I'm using Calibre 3.48 on macOS 10.13.8. Thanks.

[lomkiri]

Calibre prior to 5.00 is working with python 2, after version 5.00 it's python 3, that's why it doesn't works. If you upgrade Calibre, it will work. There is plenty of new features, you should do it.
If you want to keep your old version try this code:

def replace(match, number, file_name, metadata, dictionaries, data, functions, *args, **kwargs):
    start_at = 80
    n = int(match.group(1))
    if n >= start_at:
        return str(n+1) + match.group(2)
    return match.group(0)

[ugosan]

Hi lomkiri, you are fast as light
I'd like to install the new version of Calibre, unfortunately my OS doesn't support it.
I have modified the script because it was giving me "IndexError: no such group", but now it works fine:

def replace(match, number, file_name, metadata, dictionaries, data, functions, *args, **kwargs):
    start_at = 19
    n = int(match.group(1))
    if n >= start_at:
        return str(n-1)
    return match.group(0)

Thank you very much for your support.

[lomkiri]

Originally Posted by ugosan

I have modified the script because it was giving me "IndexError: no such group", but now it works fine

It depends of the search string you've used. In the one I wrote for Chouette:
class="noteref">\K(\d+)(</a>)
there were 2 capturing groups: group(1) -> (\d+) and group(2) -> (</a>)

You probably put a different string with only one capturing group, so you correctly adapted the code of the function.

Originally Posted by ugosan

Thank you very much for your support.

Most welcome :-).

[ugosan]

Originally Posted by lomkiri

It depends of the search string you've used. In the one I wrote for Chouette:
class="noteref">\K(\d+)(</a>)
there were 2 capturing groups: group(1) -> (\d+) and group(2) -> (</a>)
You probably put a different string with only one capturing group, so you correctly adapted the code of the function.

is exactly how you write, I had guessed that this was the reason, but now you have explained it, I understand better.
Thanks.
ugosan