我有两张桌子:
表组 - id(bigserial),name(varchar),mails(json)
表邮件 - id(bigserial),name(varchar)
我的小组数据
1, en-mails, [{"id" : 1}, {"id" : 2}]
2, fr-mails, [{"id" : 3}, {"id" : 4}]
我的邮件数据
1, mail1@gmail.com
2, mail2@gmail.com
3, mail3@gmail.com
4, mail4@gmail.com
我的查询:
SELECT tg.name, tm.mail
FROM groups as tg
CROSS JOIN LATERAL json_array_elements (tg.mails :: json) group_mails
LEFT OUTER JOIN mails as tm ON (group_mails ->> 'id') :: BIGINT = tm.c_id
我的结果
Array ( [name] => en-mails [mail] => mail1@gmail.com )
Array ( [name] => en-mails [mail] => mail2@gmail.com )
Array ( [name] => fr-mails [mail] => mail3@gmail.com )
Array ( [name] => fr-mails [mail] => mail4@gmail.com )
我的问题 - 查询如何返回:
Array ( [name] => en-mails [mail] => [mail1@gmail.com, mail2@gmail.com] )
Array ( [name] => fr-mails [mail] => [mail1@gmail.com, mail2@gmail.com] )
提前致谢
答案 0 :(得分:0)
使用聚合函数array_agg():
SELECT tg.name, array_agg(tm.mail) as mail
FROM groups as tg
CROSS JOIN LATERAL json_array_elements (tg.mails :: json) group_mails
LEFT OUTER JOIN mails as tm ON (group_mails ->> 'id') :: BIGINT = tm.id
GROUP BY 1
name | mail
----------+-----------------------------------
en-mails | {mail1@gmail.com,mail2@gmail.com}
fr-mails | {mail3@gmail.com,mail4@gmail.com}
(2 rows)