在SQL中返回JSON对象数组(Postgres)

Question

我有下表MyTable:

 id ? value_two ? value_three ? value_four 
???????????????????????????????????????????
  1 ? a         ? A           ? AA
  2 ? a         ? A2          ? AA2
  3 ? b         ? A3          ? AA3
  4 ? a         ? A4          ? AA4
  5 ? b         ? A5          ? AA5

我想查询{ value_three, value_four }按组分组的对象数组value_two.value_two应该在结果中独立存在.结果应如下所示:

 value_two ?                                                                                    value_four                                                                                 
???????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????
 a         ? [{"value_three":"A","value_four":"AA"}, {"value_three":"A2","value_four":"AA2"}, {"value_three":"A4","value_four":"AA4"}]
 b         ? [{"value_three":"A3","value_four":"AA3"}, {"value_three":"A5","value_four":"AA5"}]

它是否使用json_agg()或无关紧要array_agg().

然而,我能做的最好的事情是:

with MyCTE as ( select value_two, value_three, value_four from MyTable ) 
select value_two, json_agg(row_to_json(MyCTE)) value_four 
from MyCTE 
group by value_two;

哪个回报:

 value_two ?                                                                                    value_four                                                                                 
???????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????
 a         ? [{"value_two":"a","value_three":"A","value_four":"AA"}, {"value_two":"a","value_three":"A2","value_four":"AA2"}, {"value_two":"a","value_three":"A4","value_four":"AA4"}]
 b         ? [{"value_two":"b","value_three":"A3","value_four":"AA3"}, {"value_two":"b","value_three":"A5","value_four":"AA5"}]

value_two在对象中有一个额外的键,我想摆脱它.我应该使用哪个SQL(Postgres)查询？

Answer 1

json_build_object() 在Postgres 9.4或更新版本中

SELECT value_two, json_agg(json_build_object('value_three', value_three
                                           , 'value_four' , value_four)) AS value_four
FROM   mytable 
GROUP  BY value_two;

手册:

从可变参数列表构建JSON对象.按照惯例,参数列表由交替的键和值组成.

任何版本(包括Postgres 9.3)

row_to_json()用ROW表达式可以解决问题:

SELECT value_two
     , json_agg(row_to_json((value_three, value_four))) AS value_four
FROM   mytable
GROUP  BY value_two;

但是你松散了原始的列名.对已注册的行类型进行强制转换可避免这种情况.(临时表的行类型也用于即席查询.)

CREATE TYPE foo AS (value_three text, value_four text);  -- once in the same session

SELECT value_two
     , json_agg(row_to_json((value_three, value_four)::foo)) AS value_four
FROM   mytable
GROUP  BY value_two;

或者使用子选择而不是ROW表达式.更详细,但没有类型转换:

SELECT value_two
     , json_agg(row_to_json((SELECT t FROM (SELECT value_three, value_four) t))) AS value_four
FROM   mytable
GROUP  BY value_two;

克雷格的相关答案中有更多解释:

• 具有嵌套连接的PostgreSQL 9.2 row_to_json()

db <> 在这里摆弄
老SQL小提琴.