题解 | #平均播放进度大于60%的视频类别#

select tag, concat(round(avg(if(timestampdiff(second,start_time, end_time) >= duration,1,timestampdiff(second, start_time, end_time)/duration)*100),2),'%') as avg_play_progress from tb_user_video_log as vl
left join tb_video_info as vi on vl.video_id = vi.video_id
group by tag
having substring_index(avg_play_progress,'%',1) > 60
order by substring_index(avg_play_progress,'.',1) desc

自己的非常蠢的办法，因为需要的字符串用文本拼接成的，最后条件用substring_index和60比