Question

数据库看起来像：

ID | volume | timestamp (timestamp without time zone)
 1 | 300    | 2015-05-27 00:
 1 | 250    | 2015-05-28 00:
 2 | 13     | 2015-05-25 00:
 1 | 500    | 2015-06-28 22:
 1 | 100    | 2015-06-28 23:
 2 | 11     | 2015-06-28 21:
 2 | 15     | 2015-06-28 23:

有没有办法将每小时的价格历史合并到每天最早的1个月并将它们放回桌面？这意味着将每小时记录合并为1条记录，总金额和时间戳为00小时（我的意思是只有一天，2013-08-15 00:00:00）。

所以，想要结果：

ID | volume | timestamp
 1 | 300    | 2015-05-27 00:
 1 | 250    | 2015-05-28 00:
 2 | 13     | 2015-05-25 00:
 1 | 600    | 2015-06-28 00:
 2 | 26     | 2015-06-28 00:

Answer 1

根据没有时间的日期看起来像一个简单的组：

select id,
       sum(volume) as volume, 
       timestamp::date as timestamp
from the_table
group by id, timestamp::date
order by id, timestamp::date;

timestamp::date会将名为timestamp的列（btw是列的可怕的名称）强制转换为date，从而删除时间部分时间戳（数据类型）。

timestamp::date具体是Postgres。 ANSI SQL等价物是cast(timestamp as date)（我是否提到 timestamp 是列的可怕名称？）

Answer 2

由于你想用这个新数据修改表，我想到了一个select into。以下是我的进展（见sql fiddle here）：

选择所需数据并将其插入临时表（foo_temp）
清空你的第一张桌子（foo）
将数据从临时表复制到（现在）空表（从foo_temp到foo）
删除临时表

以下是代码：

/*Step 1 : Select the data you need and insert it in a temporary table*/
SELECT 
    ID,
    SUM(volume) as volume, 
    timestamp_field::date
INTO foo_temp
FROM foo
GROUP BY 
    ID,
    timestamp_field::date
ORDER BY 
    ID,
    timestamp_field::date;

/*Step 2 : Delete data from the table*/
DELETE FROM foo;

/*Step3 : Take data from the temporary table and insert it into the "main" table*/
INSERT INTO foo(ID,volume,timestamp_field)
SELECT * FROM foo_temp;

/*Step 4: Drop the temporary table*/
DROP TABLE foo_temp;

/*Step 5 : Yeah it works !*/
SELECT * FROM foo;

我必须承认@a_horse_with_no_name做了大部分工作，他的回答很优雅。

注意：可能有更好的方法来完成这项工作。

Answer 3

感谢您的帮助@Pholochtairze和@a_horse_with_no_name 最终查询：

WITH merged_history AS (
    SELECT item_id, SUM(history_volume) AS history_volume,
        (SUM(history_medianprice * history_volume) / SUM(history_volume)) AS history_medianprice,
        history_timestamp::date
    FROM prices_history
    WHERE history_timestamp < NOW()::date - INTERVAL '1 month'
    GROUP BY item_id, history_timestamp::date),
upsert AS ( 
    UPDATE prices_history AS h
    SET history_volume = mh.history_volume, history_medianprice = mh.history_medianprice
    FROM merged_history AS mh
    WHERE h.item_id = mh.item_id AND h.history_timestamp = mh.history_timestamp RETURNING h.*)
INSERT INTO prices_history
SELECT item_id, history_volume, history_medianprice, history_timestamp
FROM merged_history
WHERE NOT EXISTS (
    SELECT 1 FROM upsert AS u, merged_history AS mh
    WHERE u.item_id = mh.item_id AND u.history_timestamp = mh.history_timestamp);

DELETE FROM prices_history
WHERE history_timestamp != history_timestamp::date
    AND history_timestamp < NOW()::date - INTERVAL '1 month';

（完成此挑战需要2分钟（3米行）：D我将每周运行一次。稍后会稍微修改一下，因为不需要合并已合并的数据）

将每小时的价格历史合并到每日

3 个答案: