首先,我在Oracle中有一个返回阻塞的过程。这个clob包含一个json字符串,我用普利森从一个select中创建了这个字符串。如下所示:
procedure xyz
(
o_json out clob
)
is
m_json_list json_list := json_list;
m_json_temp json;
begin
for cs in (select id, name, birthday from profile)loop
m_json_temp := json;
m_json_temp.put('id', cs.id);
m_json_temp.put('name', cs.name);
m_json_temp.put('birthday', cs.birthday);
m_json_list.add(m_json_temp);
end loop;
o_json := convertToClob(m_json_list);
end xyz;现在,我想通过Postgres数据库实现同样的结果。我发现的唯一原因是我有一张有一个cloumn的桌子,它的类型是' json‘,包含了整个json。这不是我要找的。
有人能给我举个例子,说明如何在postgresql中实现这个场景吗?
编辑:这里的是一个内部连接的示例:
procedure xyz
(
o_json out clob
)
is
m_json_list json_list := json_list;
m_json_temp json;
begin
for cs in (select ppf.id, ppf.name, ppf.birthday, ott.info from profile ppf inner join other_table ott on ott.ott_id = ppf.id )loop
m_json_temp := json;
m_json_temp.put('id', cs.id);
m_json_temp.put('name', cs.name);
m_json_temp.put('birthday', cs.birthday);
m_json_temp.put('info', cs.info);
m_json_list.add(m_json_temp);
end loop;
o_json := convertToClob(m_json_list);
end xyz;发布于 2016-03-29 20:09:37
因此,您正在寻找一种从查询构造json数组的方法。
给出了一个表和一些测试数据:
postgres=# create table profile(id serial, name text, birthday date);
CREATE TABLE
postgres=# insert into profile(name, birthday) values('John', current_date - interval '30 years');
INSERT 0 1
postgres=# insert into profile(name, birthday) values('Jack', current_date - interval '25 years');
INSERT 0 1可以将行表示为json对象,如下所示:
postgres=# select row_to_json(p.*) from profile p;
row_to_json
------------------------------------------------
{"id":1,"name":"John","birthday":"1986-03-29"}
{"id":2,"name":"Jack","birthday":"1991-03-29"}
(2 rows)然后将这些json对象聚合到一个数组中:
postgres=# select json_agg(row_to_json(p.*)) from profile p;
json_agg
--------------------------------------------------------------------------------------------------
[{"id":1,"name":"John","birthday":"1986-03-29"}, {"id":2,"name":"Jack","birthday":"1991-03-29"}]
(1 row)更简单的是,您可以只使用聚合,它将为您完成所有的转换:
postgres=# select json_agg(p.*) from profile p;
json_agg
---------------------------------------------------
[{"id":1,"name":"John","birthday":"1986-03-29"}, +
{"id":2,"name":"Jack","birthday":"1991-03-29"}]
(1 row)(别管+符号了,它不是json的一部分。)
https://stackoverflow.com/questions/36293057
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