create table student
id int primary key,
name varchar(11),
course varchar(11),
score int请用一条sql查出 >80分的科目列表，且用一行记录展示
SELECT
name AS studentName,
JSON_ARRAYAGG(course) AS courseNameList
FROM student
WHERE score > 80
GROUP BY studentName;https://database.guide/json_arrayagg-create-a-json-array-from-the-rows-of-a-query-in-mysql/
原文：https://blog.51cto.com/l0vesql/2492159