postgresql组列到json包含数组
表_名称epostgresql组列到json包含数组,json,postgresql,Json,Postgresql,表_名称e id | name | cate | link ---+------+------+--------- 1 | a | A | link1 2 | a | B | link2 3 | a | B | link3 4 | b | B | link4 5 | c | A | link5 6 | d | A | link6 7 | e | B | link7 我想要回复: name |
id | name | cate | link
---+------+------+---------
1 | a | A | link1
2 | a | B | link2
3 | a | B | link3
4 | b | B | link4
5 | c | A | link5
6 | d | A | link6
7 | e | B | link7
我想要回复:
name | A | B
------+------+-------------------+------------
a | {id: 1, link: 'link1'} | [{id: 2, link: 'link2'}, {id: 3, link: 'link3'}]
b | | [{id: 4, link: 'link4'}]
c | {id: 5, link: 'link5'} |
d | {id: 6, link: 'link6'} |
e | | [{id: 7, link: 'link7'}]
cate字段值仅为A、B、C、D。但值为B必须是数组
我的实验失败了
select name, format('{%s}', string_agg(format('"id": "%s", "name": "%s", "link":"%s"', id, name, link), ','))::json as A from elements where cate = 'A' group by name;
select name, string_to_array(format('[link: "%s", id: "%s", name: "%s"]', link, id, name)) as B from elements where cate = 'B' group by name;
select name, format('{%s}', string_agg(format('"id": "%s", "name": "%s", "link":"%s"', id, name, link), ','))::json as C from elements where cate = 'C' group by name;
select name, format('{%s}', string_agg(format('"id": "%s", "name": "%s", "link":"%s"', id, name, link), ','))::json as D from elements where cate = 'D' group by name;
如果连接其他表p
id | e_id | role_id
---+------+----------
1 | 1 | 100
2 | 3 | 101
3 | 4 | 102
4 | 5 | 103
结果:
name | checked | A | B
-------+---------+---------------------------------------+-----------------------------------------------------------------
a | true | {id: 1, link: 'link1', checked: true} | [{id: 2, link: 'link2'}, {id: 3, link: 'link3', checked: true}]
b | true | | [{id: 4, link: 'link4', checked: true}]
c | true | {id: 5, link: 'link5', checked: true} |
d | | {id: 6, link: 'link6'} |
e | | | [{id: 7, link: 'link7'}]
首先,不要使用format函数来创建json字段,PostgreSQL拥有解析和创建json字段所需的所有函数。您没有发布示例ddl和数据,但您的查询应该如下所示:
select name,
json_agg(case cate when 'A' then json_build_object('id',id,'link',link) end) A,
json_agg(case cate when 'B' then json_build_object('id',id,'link',link) end) B
from data group by name
首先,不要使用format函数来创建json字段,PostgreSQL拥有解析和创建json字段所需的所有函数。您没有发布示例ddl和数据,但您的查询应该如下所示:
select name,
json_agg(case cate when 'A' then json_build_object('id',id,'link',link) end) A,
json_agg(case cate when 'B' then json_build_object('id',id,'link',link) end) B
from data group by name
你的尝试有什么问题?你有错误吗?结果不正确?如果是,哪个?postgresqlversion@JustMe9.4+我正在尝试,谢谢。你的尝试有什么问题?你有错误吗?结果不正确?如果是,哪个?postgresqlversion@JustMe9.4+我正在尝试,谢谢。字段A应该是return json。字段A应该是return json。