10

我的 PG 数据库中有一张表，看起来有点像这样：

id | widget_id | for_date | score |

每个引用的小部件都有很多这些项目。每个小部件每天 1 个，但存在差距。

我想要得到的是一个结果，其中包含自 X 以来每个日期的所有小部件。日期是通过生成系列引入的：

 SELECT date.date::date
   FROM generate_series('2012-01-01'::timestamp with time zone,'now'::text::date::timestamp with time zone, '1 day') date(date)
 ORDER BY date.date DESC;

如果给定的 widget_id 没有日期条目，我想使用前一个。所以说小部件 1337 在 2012-05-10 上没有条目，但在 2012-05-08 上没有条目，那么我希望结果集也显示 2012-05-10 上的 2012-05-08 条目：

Actual data:
widget_id | for_date   | score
1312      | 2012-05-07 | 20
1337      | 2012-05-07 | 12
1337      | 2012-05-08 | 41
1337      | 2012-05-11 | 500

Desired output based on generate series:
widget_id | for_date   | score
1336      | 2012-05-07 | 20
1337      | 2012-05-07 | 12
1336      | 2012-05-08 | 20
1337      | 2012-05-08 | 41
1336      | 2012-05-09 | 20
1337      | 2012-05-09 | 41
1336      | 2012-05-10 | 20
1337      | 2012-05-10 | 41
1336      | 2012-05-11 | 20
1337      | 2012-05-11 | 500

最终我想把它归结为一个视图，这样我每天就有一致的数据集，我可以很容易地查询。

编辑：使示例数据和预期结果集更清晰

4

4 回答 4

8

SQL小提琴

select
    widget_id,
    for_date,
    case
        when score is not null then score
        else first_value(score) over (partition by widget_id, c order by for_date)
        end score
from (
    select
        a.widget_id,
        a.for_date,
        s.score,
        count(score) over(partition by a.widget_id order by a.for_date) c
    from (
        select widget_id, g.d::date for_date
        from (
            select distinct widget_id
            from score
            ) s
            cross join
            generate_series(
                (select min(for_date) from score),
                (select max(for_date) from score),
                '1 day'
            ) g(d)
        ) a
        left join
        score s on a.widget_id = s.widget_id and a.for_date = s.for_date
) s
order by widget_id, for_date
于 2013-02-14T13:34:08.603 回答
7

首先，您可以有一个更简单的generate_series()表表达式。相当于你的（除了降序，这与你的问题的其余部分相矛盾）：

SELECT generate_series('2012-01-01'::date, now()::date, '1d')::date

该类型在输入时date被强制转换为timestamptz自动。返回类型是timestamptz任何一种方式。我在下面使用了一个子查询，因此我可以立即将其转换为输出date。

接下来，max()作为窗口函数返回您需要的内容：自框架开始忽略值以来的最高NULL值。在此基础上，您会得到一个非常简单的查询。

对于给定的 widget_id

很可能比涉及CROSS JOINor更快WITH RECURSIVE：

SELECT a.day, s.*
FROM  (
   SELECT d.day
         ,max(s.for_date) OVER (ORDER BY d.day) AS effective_date
   FROM  (
      SELECT generate_series('2012-01-01'::date, now()::date, '1d')::date
      ) d(day)
   LEFT   JOIN score s ON s.for_date = d.day
                      AND s.widget_id = 1337 -- "for a given widget_id"
   ) a
LEFT   JOIN score s ON s.for_date = a.effective_date
                   AND s.widget_id = 1337
ORDER  BY a.day;

-> sqlfiddle

使用此查询，您可以将任何score您喜欢的列放入最终SELECT列表。为了简单起见，我放了 s.*。选择你的列。

如果您想从实际有分数的第一天开始输出，只需将最后一天替换LEFT JOIN为JOIN.

所有 widget_id 的通用表单

在这里，我使用 aCROSS JOIN在每个日期为每个小部件生成一行..

SELECT a.day, a.widget_id, s.score
FROM  (
   SELECT d.day, w.widget_id
         ,max(s.for_date) OVER (PARTITION BY w.widget_id
                                ORDER BY d.day) AS effective_date
   FROM  (SELECT generate_series('2012-05-05'::date
                                ,'2012-05-15'::date, '1d')::date AS day) d
   CROSS  JOIN (SELECT DISTINCT widget_id FROM score) AS w
   LEFT   JOIN score s ON s.for_date = d.day AND s.widget_id = w.widget_id
   ) a
JOIN  score s ON s.for_date = a.effective_date
             AND s.widget_id = a.widget_id  -- instead of LEFT JOIN
ORDER BY a.day, a.widget_id;

-> sqlfiddle

于 2013-02-14T14:51:05.523 回答
2

使用您的表结构，我创建了以下递归 CTE，它以您的 MIN(For_Date) 开始并递增，直到达到 MAX(For_Date)。不确定是否有更有效的方法，但这似乎运作良好：

WITH RECURSIVE nodes_cte(widgetid, for_date, score) AS (
-- First Widget Using Min Date
 SELECT 
    w.widgetId, 
    w.for_date, 
    w.score
 FROM widgets w 
  INNER JOIN ( 
      SELECT widgetId, Min(for_date) min_for_date
      FROM widgets
      GROUP BY widgetId
   ) minW ON w.widgetId = minW.widgetid 
        AND w.for_date = minW.min_for_date
UNION ALL
 SELECT 
    n.widgetId,
    n.for_date + 1 for_date,
    coalesce(w.score,n.score) score
 FROM nodes_cte n
  INNER JOIN (
      SELECT widgetId, Max(for_date) max_for_date
      FROM widgets 
      GROUP BY widgetId
   ) maxW ON n.widgetId = maxW.widgetId
  LEFT JOIN widgets w ON n.widgetid = w.widgetid 
    AND n.for_date + 1 = w.for_date
  WHERE n.for_date + 1 <= maxW.max_for_date
)
SELECT * 
FROM nodes_cte 
ORDER BY for_date

这是SQL 小提琴。

以及返回的结果（根据需要格式化日期）：

WIDGETID   FOR_DATE                     SCORE
1337       May, 07 2012 00:00:00+0000   12
1337       May, 08 2012 00:00:00+0000   41
1337       May, 09 2012 00:00:00+0000   41
1337       May, 10 2012 00:00:00+0000   41
1337       May, 11 2012 00:00:00+0000   500

请注意，这假设您的 For_Date 字段是一个日期——如果它包含一个时间——那么您可能需要在上面的查询中使用间隔“1 天”。

希望这可以帮助。

于 2013-02-14T13:23:11.067 回答
0

数据：

DROP SCHEMA tmp CASCADE;
CREATE SCHEMA tmp ;
SET search_path=tmp;

CREATE TABLE widget
        ( widget_id INTEGER NOT NULL
        , for_date DATE NOT NULL
        , score INTEGER
         , PRIMARY KEY (widget_id,for_date)
        );
INSERT INTO widget(widget_id , for_date , score) VALUES
 (1312, '2012-05-07', 20)
, (1337, '2012-05-07', 12)
, (1337, '2012-05-08', 41)
, (1337, '2012-05-11', 500)
        ;

查询：

SELECT w.widget_id AS widget_id
        , cal::date AS for_date
        -- , w.for_date AS org_date
        , w.score AS score
FROM generate_series( '2012-05-07'::timestamp , '2012-05-11'::timestamp
                 , '1day'::interval) AS cal
        -- "half cartesian" Join;
        -- will be restricted by the NOT EXISTS() below
LEFT JOIN widget w ON w.for_date <= cal
WHERE NOT EXISTS (
        SELECT * FROM widget nx
        WHERE nx.widget_id = w.widget_id
        AND nx.for_date <= cal
        AND nx.for_date > w.for_date
        )
ORDER BY cal, w.widget_id
        ;

结果：

 widget_id |  for_date  | score 
-----------+------------+-------
      1312 | 2012-05-07 |    20
      1337 | 2012-05-07 |    12
      1312 | 2012-05-08 |    20
      1337 | 2012-05-08 |    41
      1312 | 2012-05-09 |    20
      1337 | 2012-05-09 |    41
      1312 | 2012-05-10 |    20
      1337 | 2012-05-10 |    41
      1312 | 2012-05-11 |    20
      1337 | 2012-05-11 |   500
(10 rows)
于 2013-02-14T19:42:44.317 回答