小编Gau*_*uss的帖子

使用 Postgres：

SELECT users."name" AS "name"
    , array_to_json(array_agg(sites)) as sites
FROM remodel.users AS users
JOIN remodel.user_sites AS user_sites
    ON users.id=user_sites.user
JOIN remodel.sites AS sites
    ON sites.id=user_sites.site
GROUP BY "users".id
;

目前生产

SELECT users."name" AS "name"
    , array_to_json(array_agg(sites)) as sites
FROM remodel.users AS users
JOIN remodel.user_sites AS user_sites
    ON users.id=user_sites.user
JOIN remodel.sites AS sites
    ON sites.id=user_sites.site
GROUP BY "users".id
;

但是我不想在 JSON 输出中有“id”字段。

将站点选择更改为

array_to_json(array_agg(row(sites."name", sites.created))) as sites

导致字段失去名称

"Toby";"[{"id":1,"name":"Village","created":"2015-08-10T15:22:36.622298"},
         {"id":2,"name":"Manor","created":"2015-08-10T15:22:43.614551"}]"
"Amy";"[{"id":3,"name":"Park","created":"2015-08-10T15:22:48.810872"}]"
"Anne";"[{"id":2,"name":"Manor","created":"2015-08-10T15:22:43.614551"},
         {"id":1,"name":"Village","created":"2015-08-10T15:22:36.622298"},
         {"id":3,"name":"Park","created":"2015-08-10T15:22:48.810872"}]"

并尝试（这将是一个可怕的）子选择

, array_to_json(array_agg((SELECT "name", created FROM …