所以我有这个表employee_info
:
branch_id | employee_id | first_namelast_name | |
---|---|---|---|
2 | 1 | "Barack"> | ";奥巴马 |
2 | 2 | ";Johnny"> | ";Depp"> |
2 | 3 | ";Emma"> | ";Watson"> |
4 | 4 | ";Daniel"> | ";Radcliffe"> |
4 | 5 | ";莱昂纳多 | ";迪卡普里奥 |
7 | 17 | ";Denzel"> | ";华盛顿"> |
4 | 13 | ";基努 | ";Reeves"> |
7 | 15 | ";Samuel"> | ";Jackson"> |
7 | 19 | ";Brendan"> | ";Fraser"> |
7 | 20 | ";Kanye"> | ";West"> |
通过使用JSON_AGG
和json_build_object
的组合来解决此问题
SELECT branch_id, JSON_AGG (json_build_object('employee_id',employee_id,'first_name',first_name,'last_name', last_name)) employees FROM employee_info GROUP BY branch_id
欢迎采用任何其他方法。