2

我有下表：

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

使用以下数据：

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”。

我需要找到 Jim 2012 年的所有销售额。任何以“sales_”开头的东西，然后把它放在一个对象中（只需要销售的产品是什么和价值）。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得不到。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}]
4

2 回答 2

4

您将第一个连接的结果视为 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。jsonb如果需要，可以选择转换为- 或jsonb_object_agg()在 Postgres 9.5 或更高版本中使用。

使用显式JOIN语法在最明显的位置附加条件。没有显式语法
的相同：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;
于 2015-07-22T00:52:44.533 回答
0

您的第一个查询可以这样解决（从臀部射击，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;

这里的诀窍是您使用LATERAL连接来“剥离”对象的外层jsonb以更深入地获取数据。此查询假设 Jim 可能在多个位置进行销售。

（处理您的查询 2）

于 2015-07-22T01:09:35.777 回答