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在 PostgreSQL 13 中，我有一个 table_a，其中每个时间戳都有许多行和列。我试图找到 X 列中的值最接近从另一个表获得的基准值的行。第二个表对于每个时间戳只有一个基准值。对于每个时间戳，我需要返回 table_a 的大部分列。直接提供基准值时，下面的查询可以正常工作。

如何从 table_b 获取基准值以在此查询中使用？

简单地将 table_b.benchmark 替换为 (SELECT benchmark FROM table_b WHERE table_a.timestamp = table_b.timestamp) 会导致“关系“t1”不存在”错误。

也想不出一个有效的连接。

表_a：

+-----------------+-----+---------------+
|    timestamp    |  x  | other_columns |
+-----------------+-----+---------------+
| 2020-01-01 8:00 | 100 |               |
| 2020-01-01 8:00 | 200 |               |
| 2020-01-01 8:00 | 300 |               |
| 2020-01-01 8:00 | 400 |               |
| 2020-01-01 8:00 | 500 |               |
| ...             |     |               |
| 2020-01-01 9:00 | 100 |               |
| 2020-01-01 9:00 | 200 |               |
| 2020-01-01 9:00 | 300 |               |
| 2020-01-01 9:00 | 400 |               |
| 2020-01-01 9:00 | 500 |               |
| ...             |     |               |
+-----------------+-----+---------------+

表_b：

+-----------------+-----------+
|    timestamp    | benchmark |
+-----------------+-----------+
| 2020-01-01 8:00 |       340 |
| 2020-01-01 9:00 |       380 |
| ...             |           |
+-----------------+-----------+

预期结果：

+-----------------+-----+
|    timestamp    |  x  |
+-----------------+-----+
| 2020-01-01 8:00 | 300 |
| 2020-01-01 9:00 | 400 |
| ...             |     |
+-----------------+-----+

SQL查询：

WITH date_filter AS (
    SELECT *
    FROM table_a
    WHERE timestamp >= {start_date} and timestamp < {end_date}
    )

SELECT DISTINCT t1.timestamp, t1.x, t1.etc
FROM date_filter AS t1

INNER JOIN (
    SELECT timestamp, MIN(ABS(x - (table_b.benchmark))) AS target_value
    FROM t1
    GROUP BY timestamp
    ) AS t2

ON t2.timestamp = t1.timestamp AND t2.target_value = ABS(x - (table_b.benchmark))
ORDER BY timestamp ASC;```
4

2 回答 2

0

我建议横向连接：

select b.*, a.x
from table_b b left join lateral
     (select a.*
      from table_a a
      where a.timestamp = b.timestamp
      order by abs(a.x - b.benchmark)
      limit 1
     ) b
     on 1=1;
于 2020-11-29T22:34:36.320 回答
0

一种选择使用横向连接：

select b.timestamp, a.x
from table_b b
cross join lateral (
    select a.*
    from table_a a
    where a.timestamp = b.timestamp
    order by abs(a.x - b.benchmark)
    limit 1
) a

您还可以使用distinct on：

select distinct on (b.timestamp) b.timestamp, a.x
from table_b b
inner join table_a a on a.timestamp = b.timestamp
order by b.timestamp, abs(a.x - b.benchmark)
于 2020-11-29T22:35:34.807 回答