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我在 postgress 中有以下 3 个表...

Employee (employee_id, ...)
Address (address_id, owner_id, ...)
Address_Component (address_id, key, value)

而地址组件具有类似...的数据

| 123 | province     | Toronto      |
| 123 | country      | Canada       |
| 123 | addressLine1 | some address |

员工可以有多个地址，每个地址与一个地址组件相关联。

我希望将地址作为 json 并将 address_component 作为嵌入在地址对象中的另一个 json。

select emp.employee_id,
    array_to_json(array_agg(row_to_json(adr))) as address,
    json_object(array_agg(adrcomp.key), array_agg(adrcomp.value)) as address_component
   FROM employee emp
   LEFT JOIN address adr ON adr.owner_id = emp.employee_id
   LEFT JOIN address_component adrcomp ON adr.address_id = adrcomp.address_id
   WHERE employee_id = 'a6f49ab5-1769-4953-9b0e-6c12754d33c7'
   GROUP BY emp.employee_id

有了这个，我能够正确地将地址数组作为 json 获取，而我暂时将地址组件作为单独的结果。你能帮我把这个 address_component json 合并到 address json 数组的单个元素中，得到如下所示的输出吗？

[{ address_id:"123", owner_id: "E1", address_component: {province:"", "country":"", addressLine1:""} }, { ... }]
4

1 回答 1

0

您不能在同一级别上做所有事情，因为您要进行两次聚合。您需要address_component从它自己的子查询中创建：

SELECT emp.employee_id,
       jsonb_agg((
         SELECT a FROM (
           SELECT adr.address_id, adr.owner_id,
                  (SELECT json_object_agg(key, value)
                   FROM address_component 
                   WHERE address_id = adr.address_id) AS address_component) a))
FROM employee emp
LEFT JOIN address adr ON adr.owner_id = emp.employee_id
WHERE employee_id = 'a6f49ab5-1769-4953-9b0e-6c12754d33c7'
GROUP BY emp.employee_id
于 2016-07-01T02:50:38.200 回答