Given a set of URL in the format {FIXED PREFIX}/{NUMERIC ID}/..., how to retrieve the URL with the highest {NUMERIC ID}?

I want to make a query that contains the Artist Name and the respective cover video, which is the highest ID on the Link Column. In this example, the second row with ..embd/125/... The link format never changes. I am using PHP ,MySQL and innodb engine

I started by doing an INNER JOIN on these two tables:

Id | Artist
1    John Doe
2    Jane Doe

And this one

Id | Id_Artist | Link 
53   1           http://example.com/embd/123/video/...
54   1           http://example.com/embd/125/video/...
55   1           http://example.com/embd/124/video/...   
56   1           http://example.com/embd/122/video/... 
57   2           http://example.com/embd/128/video/...
58   2           http://example.com/embd/129/video/...    
59   2           http://example.com/embd/130/video/...

Which outputs the expected result with all the artists and they're videos

SELECT fp.id, fp.name, gl.link AS Video
FROM feed_post_artist AS fp
INNER JOIN gallery gl ON gl.id_artist = fp.id

Id | Id_Artist | Link 
 1   John Doe    http://example.com/embd/123/video/...
 1   John Doe    http://example.com/embd/125/video/...
 1   John Doe    http://example.com/embd/124/video/...   
 1   John Doe    http://example.com/embd/122/video/... 
 2   Jane Doe    http://example.com/embd/128/video/...
 2   Jane Doe    http://example.com/embd/129/video/...    
 2   Jane Doe    http://example.com/embd/130/video/...

Is it possible to use regex to compare each artist video and return only the highest link id as a result? ( First 4 rows 125 is highest, last 3 rows 130 is the highest ). I have tried using MAX() but it won't work, obviously.

Id | Id_Artist | Link 
 1   John Doe    http://example.com/embd/125/video/...
 2   Jane Doe    http://example.com/embd/130/video/...

EDIT :

As Julian in the comments suggested, MySQL SUBSTR() solves this particular case. However, when the id goes over 999, it will stop working.

SELECT fp.id, fp.name, SUBSTR(gl.link,28,5) AS Video
FROM feed_post_artist AS fp
INNER JOIN gallery gl ON gl.id_artist = fp.id

The only part missing is grouping the results to return only the MAX()

Solution

select      a.Id
           ,a.Artist
           ,g.Link

from                    feed_post_artist    as a

            left join  (select      Id_Artist
                                   ,substr(max(concat(lpad(substring_index(substring_index(Link,'/',5),'/',-1),10,'0'),Link)),11)   as Link  

                        from        gallery as g

                        group by    Id_Artist
                        ) g

            on          g.Id_Artist =
                        a.Id

order by    a.Id

+----+----------+---------------------------------------+
| Id | Artist   | Link                                  |
+----+----------+---------------------------------------+
| 1  | John Doe | http://example.com/embd/125/video/... |
+----+----------+---------------------------------------+
| 2  | Jane Doe | http://example.com/embd/130/video/... |
+----+----------+---------------------------------------+