如何从我的数据返回jsonb数组和对象数组？

一尘不染

如何从我的数据返回jsonb数组和对象数组？

sql

我有下表：

CREATE TABLE mytable (
  id       serial PRIMARY KEY
, employee text UNIQUE NOT NULL
, data     jsonb
);

带有以下数据：

INSERT INTO mytable (employee, data)
VALUES
 ('Jim', '{"sales_tv": [{"value": 10, "yr": "2010", "loc": "us"}, {"value": 5, "yr": "2011", "loc": "europe"}, {"value": 40, "yr": "2012", "loc": "asia"}], "sales_radio": [{"value": 11, "yr": "2010", "loc": "us"}, {"value": 8, "yr": "2011", "loc": "china"}, {"value": 76, "yr": "2012", "loc": "us"}], "another_key": "another value"}'),
 ('Rob', '{"sales_radio": [{"value": 7, "yr": "2014", "loc": "japan"}, {"value": 3, "yr": "2009", "loc": "us"}, {"value": 37, "yr": "2011", "loc": "us"}], "sales_tv": [{"value": 4, "yr": "2010", "loc": "us"}, {"value": 18, "yr": "2011", "loc": "europe"}, {"value": 28, "yr": "2012", "loc": "asia"}], "another_key": "another value"}')

请注意，除了“ sales_tv”和“ sales_radio”之外，还有其他键。对于下面的查询，我只需要关注“ sales_tv”和“
sales_radio”。

我需要查找Jim在2012年的所有销售额。任何以“ sales_”开头的内容，然后将其放在一个对象中（只需要所售产品的价值和价值）即可。例如：

    employee   | sales_
    Jim        | {"sales_tv": 40, "sales_radio": 76}

我有：

SELECT * FROM mytable,
  (SELECT l.key, l.value FROM mytable, lateral jsonb_each_text(data) AS l
    WHERE key LIKE 'sales_%') AS a,
  jsonb_to_recordset(a.value::jsonb) AS d(yr text, value float)
  WHERE mytable.employee = 'Jim'
  AND d.yr = '2012'

但我似乎甚至无法获得吉姆的数据。相反，我得到：

employee | key         |  value
-------- |------       | -----
Jim      | sales_tv    |  [{"yr": "2010", "loc": "us", "value": 4}, {"yr": "2011", "loc": "europe", "value": 18}, {"yr": "2012", "loc": "asia", "value": 28}]
Jim      | sales_tv    |  [{"yr": "2010", "loc": "us", "value": 10}, {"yr": "2011", "loc": "europe", "value": 5}, {"yr": "2012", "loc": "asia", "value": 40}]
Jim      | sales_radio |  [{"yr": "2010", "loc": "us", "value": 11}, {"yr": "2011", "loc": "china", "value": 8}, {"yr": "2012", "loc": "us", "value": 76}]

阅读 167

2021-03-17

共1个答案

一尘不染

您将第一次连接的结果视为JSON，而不是文本字符串，因此请使用
jsonb_each() 代替jsonb_each_text()：

SELECT t.employee, json_object_agg(a.k, d.value) AS sales
FROM   mytable t
JOIN   LATERAL jsonb_each(t.data) a(k,v) ON a.k LIKE 'sales_%'
JOIN   LATERAL jsonb_to_recordset(a.v) d(yr text, value float) ON d.yr = '2012'
WHERE  t.employee = 'Jim'  -- works because employee is unique
GROUP  BY 1;

GROUP BY 1是的简写GROUP BY t.employee。
结果：

employee | sales
---------+--------
Jim      | '{ "sales_tv" : 40, "sales_radio" : 76 }'

我还理清和简化了您的查询。

json_object_agg()
在将名称/值对作为JSON对象进行聚合时发挥了作用。jsonb如果需要，可以选择强制转换为-或jsonb_object_agg()在Postgres 9.5或更高版本中使用。

使用显式JOIN语法将条件附加在最明显的位置。没有显式语法
的情况相同JOIN：

SELECT t.employee, json_object_agg(a.k, d.value) AS sales
FROM   mytable t
     , jsonb_each(t.data)      a(k,v) 
     , jsonb_to_recordset(a.v) d(yr text, value float)
WHERE  t.employee = 'Jim'
AND    a.k LIKE 'sales_%'
AND    d.yr = '2012'
GROUP  BY 1;

2021-03-17