IN查询排序:
select * from test where id in(3,1,5) order by find_in_set(id,’3,1,5′);
select * from test where id in(3,1,5) order by substring_index(‘3,1,2’,id,1);
正则使用:
SELECT COUNT(alarmID)
FROM Alarm
WHERE (CVE NOT RLIKE ‘^CVE-[0-9]{4}-[0-9]{4}$’ OR CVE IS NULL)