有如下表

其中awardee和awardee_unit都是保存的json类型的字符串, awardee是多个人员id, awardee_unit是部门的全路径

查询时要注意转换 

需要将name拼接起来合并成一行,直接 GROUP_CONCAT 会报错

百度的大部分答案是修改数据库配置去掉严格模式,如果不方便修改数据库可以这样做

select a.id,a.award_name,a.project_name,d.dept_name,a.awardee_time,
(SELECT GROUP_CONCAT(c.name) FROM biz_employee c WHERE FIND_IN_SET(c.id,REPLACE(SUBSTRING(a.awardee,2, LENGTH(awardee)-2), ' ', ''))) awardee
from biz_technology_awards a
join biz_technology_awards_emp b on a.id = b.technology_awards_id
join biz_employee c on b.emp_id = c.id
join sys_dept d on d.dept_id = JSON_EXTRACT(a.awardee_unit, concat('$[', json_length(a.awardee_unit) - 1, ']'))
where c.id = 1