如何从我的数据中 return 一个 jsonb 数组和对象数组？

Question

我有以下 table:

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

具有以下数据：

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。以 "sales_" 开头的所有内容，然后将其放入一个对象中（只需要所售产品的数量和价值）。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'

但我似乎连吉姆的数据都无法获取。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}]

Answer 1

您将第一个连接的结果视为 JSON，而不是文本字符串，因此使用 jsonb_each() 而不是 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 shorthand 对于 GROUP BY t.employee。
结果：

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

我还整理并简化了您的查询。

json_object_agg() is instrumental in aggregating name/value pairs as JSON object. Optionally cast to jsonb if you need that - or use jsonb_object_agg() 在 Postgres 9.5 或更高版本中。

使用显式 JOIN 语法在最明显的位置附加条件。
没有显式 JOIN 语法的相同：

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;

Answer 2

你的第一个问题可以这样解决（偷懒，这里无法访问PG 9.4）：

SELECT employee, json_object_agg(key, sales)::jsonb AS sales_
FROM (
  SELECT t.employee, j.key, sum((e->>'value')::int) AS sales
  FROM mytable t,
       jsonb_each(t.data) j,
       jsonb_array_elements(j.value) e
  WHERE t.employee = 'Jim'
    AND j.key like 'sales_%'
    AND e->>'yr' = '2012'
  GROUP BY t.employee, j.key) sub
GROUP BY employee;

这里的技巧是您使用 LATERAL 连接到 jsonb 对象的 "peel away" 外层以更深入地获取数据。此查询假设 Jim 可能在多个地点有销售。

（处理您的查询 2）

如何从我的数据中 return 一个 jsonb 数组和对象数组？

How do I return a jsonb array and array of objects from my data?

sql

postgresql

postgresql-9.4

jsonb