题解 | 提取博客URL中的用户名

select
    device_id,
    substring_index(blog_url, '/', -1) as user_name
from
    user_submit

select
    device_id, replace(blog_url,'http:/url/','') as user_name
from
    user_submit;

select
    device_id, substr(blog_url,11,length(blog_url)-10) as user_nam
from
    user_submit;

select
    device_id, trim('http:/url/' from blog_url) as user_name
from
    user_submit;

🌹结束！睡觉！！！！