0

现在我尝试解决一个问题，我的理解似乎与在巨大的事件流中寻找差距有关？

我的表中有几个数据流。我想按时间总结它们，但它们并不总是相同的时间戳。该表如下所示：

架构：

CREATE TABLE Table1
    ("id" int, "stream_id" int, "timestamp" timestamp, "value" real)
;

INSERT INTO Table1
    ("id", "stream_id", "timestamp", "value")
VALUES
    (1, 7, '2015-06-01 15:20:30', 0.1),
    (2, 7, '2015-06-01 15:20:31', 0.2),
    (3, 7, '2015-06-01 15:20:32', 0.3),
    (4, 7, '2015-06-01 15:25:30', 0.5),
    (5, 7, '2015-06-01 15:25:31', 1.0),

    (6, 6, '2015-06-01 15:20:31', 1.1),
    (7, 6, '2015-06-01 15:20:32', 1.2),
    (8, 6, '2015-06-01 15:20:33', 1.3),
    (9, 6, '2015-06-01 15:25:31', 1.5),
    (10, 6, '2015-06-01 15:25:32', 2.0)    
;

我尝试解决它：

with ts as (select "timestamp"       
           from Table1
           order by "timestamp"
          ),
data as (select "timestamp","value"
         from Table1
         order by "timestamp"
         ),
streams as (select "stream_id"        
           from Table1
           group by "stream_id"
           order by "stream_id"
          )          
select * .... (question)

我希望得到所有汇总数据的图表线。当一次，在其他流中没有数据时，总和应该取行，timestamp < current_timestamp但在当前 time_stamp 处最近。如果没有值，则假定为 0。

我考虑过递归查询，但不知何故我看不到解决方案......

编辑：在这里我试图以图形方式解释它：

我的问题的图形描述

编辑2：

我考虑过这样的事情，但我没有得到最后一个“东西”来完成它。

with RECURSIVE data as (
    select * from rawdata 
    where date(date_time)='2014-05-01'
),
streams as (
    select stream_id from data 
    group by stream_id
),
t(n) AS (
    VALUES (1)
  UNION ALL
    SELECT n+1 FROM t WHERE n < (select count(*) from streams)
)
SELECT n FROM t;
4

1 回答 1

1

很抱歉，上一个查询中有错误
这是一个新的更正查询：

WITH times AS(
  SELECT DISTINCT "timestamp" As tm
  FROM table1
)
SELECT tm, SUM( val ) as s_u_m
FROM (
    SELECT tm, "stream_id",
           (  SELECT "value" FROM Table1 t2
              WHERE t2."timestamp" = max( t1."timestamp" )
                AND t2."stream_id" = t1."stream_id"
              ORDER BY "id" DESC LIMIT 1
            ) As val
    FROM times t
    JOIN table1 t1
    ON t.tm >= t1."timestamp"
    GROUP BY tm, "stream_id"
    order by tm
  ) you_must_have_an_alias_here_in_order_to_avoid_the_syntax_error
GROUP BY tm
ORDER BY tm;
;

以及源数据中包含 3 个流的演示：http ://sqlfiddle.com/#!15/30eb8/5

这是一个具有模仿图形布局的布局的源表：

|  x | id |              timestamp | stream6 | stream7 | stream8 |
|----|----|------------------------|---------|---------|---------|
|  1 |  1 | June, 01 2015 15:20:30 |  (null) |     0.1 |  (null) |
|  2 |  2 | June, 01 2015 15:20:31 |  (null) |     0.2 |  (null) |
|  3 |  3 | June, 01 2015 15:20:31 |     1.1 |  (null) |  (null) |
|  4 |  4 | June, 01 2015 15:20:32 |  (null) |     0.3 |  (null) |
|  5 |  5 | June, 01 2015 15:20:32 |     1.2 |  (null) |  (null) |
|  6 | 11 | June, 01 2015 15:20:32 |  (null) |  (null) |     2.3 |
|  7 | 12 | June, 01 2015 15:20:32 |  (null) |  (null) |     1.1 |
|  8 | 10 | June, 01 2015 15:20:33 |     1.3 |  (null) |  (null) |
|  9 | 13 | June, 01 2015 15:20:33 |  (null) |  (null) |     1.7 |
| 10 |  6 | June, 01 2015 15:25:30 |  (null) |     0.5 |  (null) |
| 11 |  7 | June, 01 2015 15:25:31 |     1.5 |  (null) |  (null) |
| 12 |  8 | June, 01 2015 15:25:31 |  (null) |       1 |  (null) |
| 13 |  9 | June, 01 2015 15:25:32 |       2 |  (null) |  (null) |

结果是：（ v(3) 表示：来自 x=3 的记录的值）

|                     tm |     s_u_m |
|------------------------|-----------|
| June, 01 2015 15:20:30 |       0.1 |  0    + v(1) + 0
| June, 01 2015 15:20:31 | 1.3000001 |  v(3) + v(2)  + 0
| June, 01 2015 15:20:32 |       2.6 |  v(5) + v(4) + v(7) => see note below !!!
| June, 01 2015 15:20:33 |       3.3 |  v(8) + v(4) + v(9)
| June, 01 2015 15:25:30 |       3.5 |  v(8) + v(10)+ v(9)
| June, 01 2015 15:25:31 |       4.2 |  v(11)+ v(12)+ v(9)
| June, 01 2015 15:25:32 |       4.7 |  v(13)+ v(12)+ v(9)

记录注释 | June, 01 2015 15:20:32 | 2.6 |

演示中的源表包含两条日期相同且source_id相同的记录：

|  6 | 11 | June, 01 2015 15:20:32 |  (null) |  (null) |     2.3 |
|  7 | 12 | June, 01 2015 15:20:32 |  (null) |  (null) |     1.1 |

ORDER BY "id" DESC由于在此代码片段中，查询仅获取最新记录 x=7 ：

(  SELECT "value" FROM Table1 t2
      WHERE t2."timestamp" = max( t1."timestamp" )
        AND t2."stream_id" = t1."stream_id"
      ORDER BY "id" DESC LIMIT 1
    ) As val

如果您想获取第一条记录 x=6 而不是最新的，DESC则从 order by 子句中删除。

如果您想对具有相同日期和 stream_id 的所有记录求和（在上面的示例中 - 记录 6 + 7），则将上述查询更改为：

(  SELECT SUM("value") FROM Table1 t2
      WHERE t2."timestamp" = max( t1."timestamp" )
        AND t2."stream_id" = t1."stream_id"
    ) As val

如果您想获取随机记录，请使用ORDER BY random().

于 2015-06-03T17:04:03.557 回答