Question

我有下表：

rpart.control

使用以下数据：

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

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”。

我需要找到2012年吉姆的所有销售。任何以“sales_”开头然后把它放在一个对象中的东西（只需要产品销售的是什么和价值）。 e.g：

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"}')

我有：

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

但我似乎无法得到吉姆的数据。相反，我得到：

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'

Answer 1

您将第一次加入的结果视为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GROUP BY t.employee的简写结果：

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

我也解开并简化了您的查询。

json_object_agg()有助于将名称/值对聚合为JSON对象。如果需要，可以选择转换为jsonb - 或者在Postgres 9.5或更高版本中使用jsonb_object_agg()。

使用明确的JOIN语法将条件附加到最明显的位置 相同的，没有明确的JOIN语法：

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

您的第一个查询可以像这样解决（从臀部拍摄，此处无法访问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;

这里的诀窍是你使用LATERAL加入＆＃34;剥离＆＃34; jsonb对象的外层，以获取更深层次的数据。此查询假设Jim可能在多个位置进行销售。

（处理您的查询2）

如何从数据中返回jsonb数组和对象数组？

2 个答案: