[英]SELECT JSON_ARRAYAGG FROM TABLE name as parameter problem
[英]Mysql JSON_ARRAYAGG from another table in select
我有职位空缺和部门,我想将所有职位空缺的部门作为 json 数组。 我尝试使用 JSON_ARRAYAGG 函数,但出现 mysql 语法错误。
select vd.id, vd.title, vd.sort,
(select json_arrayagg(v.id, v.title) from vacancies as v where vd.id = v.department_id) as vacancies
from vacancy_departments as vd
如果您提供了一些样本数据和预期结果,我们可以为您提供更好的服务。 同时尝试这样的事情:
SELECT vd.id, vd.title, vd.sort,
(
SELECT JSON_ARRAYAGG(
JSON_MERGE_PRESERVE(
JSON_OBJECT("id", v.id),
JSON_OBJECT("title", v.title)
))
FROM vacancies as v
WHERE vd.id = v.department_id
) as vacancies
FROM vacancy_departments as vd
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