您可以直接从 MySQL 生成 JSON 内容。这是一个适用于 MySQL 5.7 或更高版本的解决方案。
作为初学者,请考虑功能JSON_OBJECT() https://dev.mysql.com/doc/refman/8.0/en/json-creation-functions.html#function_json-object,为表中的每条记录生成一个 JSON 对象:
SELECT
p.*,
JSON_OBJECT('id', id, 'project_name', project_name, 'parent_id', parent_id) js
FROM tbl_projects p;
给定您的样本数据,这将返回:
| id | project_name | parent_id | js |
| --- | ------------------- | --------- | ---------------------------------------------------------------- |
| 1 | Carmichael House | 0 | {"id": 1, "parent_id": 0, "project_name": "Carmichael House"} |
| 2 | Carmichael Kitchen | 1 | {"id": 2, "parent_id": 1, "project_name": "Carmichael Kitchen"} |
| 3 | Carmichael Bathroom | 1 | {"id": 3, "parent_id": 1, "project_name": "Carmichael Bathroom"} |
| 4 | Dowd Apartment | 0 | {"id": 4, "parent_id": 0, "project_name": "Dowd Apartment"} |
| 5 | Dowd Kitchen | 4 | {"id": 5, "parent_id": 4, "project_name": "Dowd Kitchen"} |
为了产生您预期的输出,我们将自行JOIN
表查找子记录,并使用聚合函数JSON_ARRAYAGG() https://dev.mysql.com/doc/refman/8.0/en/group-by-functions.html#function_json-arrayagg生成内部 JSON 数组。额外的聚合级别将所有内容填充到单个对象中。如示例数据所示,我假设根项目有parent_id = 0
并且只有一层层次结构:
SELECT JSON_OBJECT('projects', JSON_ARRAYAGG(js)) results
FROM (
SELECT JSON_OBJECT(
'id', p.id,
'project_name', p.project_name,
'parent_id', p.parent_id,
'children', JSON_ARRAYAGG(
JSON_OBJECT(
'id', p1.id,
'project_name', p1.project_name,
'parent_id', p1.parent_id
)
)
) js
FROM tbl_projects p
LEFT JOIN tbl_projects p1 ON p.id = p1.parent_id
WHERE p.parent_id = 0
GROUP BY p.id, p.project_name, p.parent_id
) x
Yields:
| results |
| -------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |
| {"projects": [{"id": 1, "children": [{"id": 2, "parent_id": 1, "project_name": "Carmichael Kitchen"}, {"id": 3, "parent_id": 1, "project_name": "Carmichael Bathroom"}], "parent_id": 0, "project_name": "Carmichael House"}, {"id": 4, "children": [{"id": 5, "parent_id": 4, "project_name": "Dowd Kitchen"}], "parent_id": 0, "project_name": "Dowd Apartment"}]} |
DB Fiddle 上的演示 https://www.db-fiddle.com/f/mr81oyYKLHCdoRB88TErS3/0