Question

要映射的数组（名称，值）。

[{'name':'email,'value':'email@email.em'},{'name':'phone','value':123123}]

到

{'email':'email@email.em','phone':123123}

我需要组织搜索和排序此类电子邮件，我想简化以View使用地图数据结构Table。

Answer 1

我将您的输入数据更改为实际上是json（dbl引用而不是单个并添加了数组的键名。此后转换是如何“构建”它的示例：

t=# with j as (
  select '{"arr":[{"name":"email","value":"email@email.em"},{"name":"phone","value":123123}]}'::json v
)
select
  concat('{',v->'arr'->0->'name',':',v->'arr'->0->'value',',',v->'arr'->1->'name',':',v->'arr'->1->'value','}')::json
from j
;
 {"email":"email@email.em","phone":123123}

Answer 2

with t(id,j) as (
  values(1,'[{"name":"email","value":"email@email.em"},{"name":"phone","value":123123}]'::json))
select json_object_agg(a.j->>'name', a.j->>'value')
from t, json_array_elements(j) a(j) group by id;
╔════════════════════════════════════════════════════╗
║                  json_object_agg                   ║
╠════════════════════════════════════════════════════╣
║ { "email" : "email@email.em", "phone" : "123123" } ║
╚════════════════════════════════════════════════════╝

如何更改PostgreSQL中JSON字段的结构？

2 个答案: