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这是我之前发布的这个出色回答的问题的变体：

我有一个数据库表：

id | date       | position | name
--------------------------------------
1  | 2016-06-29 | 9        | Ben Smith
2  | 2016-06-29 | 1        | Ben Smith
3  | 2016-06-29 | 5        | Ben Smith
4  | 2016-06-29 | 6        | Ben Smith
5  | 2016-06-30 | 2        | Ben Smith
6  | 2016-06-30 | 2        | Tom Brown
7  | 2016-06-29 | 4        | Tom Brown
8  | 2016-06-30 | 2        | Tom Brown
9  | 2016-06-30 | 1        | Tom Brown

如何有效地查询表，以便使用 array_agg() 获取新列。

我已经尝试过以下查询，但是它非常慢而且错误，因为它没有按名称列对 previous_positions 进行分组：

 SELECT runners.id AS runner_id,
    btrim(regexp_replace(replace(upper(runners.name::text), '.'::text, ''::text), '[[:digit:]]'::text, ''::text, 'g'::text)) AS name,
    runners.position_two,
    (array_agg(runners.position_two) OVER w AS results
   FROM runners
  WINDOW w AS (PARTITION BY (btrim(regexp_replace(replace(upper(runners.name::text), '.'::text, ''::text), '[[:digit:]]'::text, ''::text, 'g'::text))) ORDER BY runners.id ROWS BETWEEN UNBOUNDED PRECEDING AND 1 PRECEDING);

我希望表格输出看起来像这样

id | date       | position | name      | previous   | med  |med_20
----------------------------------------------------------------------
1  | 2016-06-29 | 9        | Ben Smith | {}         |      |
2  | 2016-06-29 | 1        | Ben Smith | {9}        | 9    | 9
3  | 2016-06-29 | 5        | Ben Smith | {9,1}      | 5    | 5
4  | 2016-06-29 | 6        | Ben Smith | {9,1,5}    | 5    | 5
5  | 2016-06-30 | 2        | Ben Smith | {9,1,5,6}  | 5.5  | 5.5
6  | 2016-06-30 | 2        | Tom Brown | {}         | None | None
7  | 2016-06-29 | 4        | Tom Brown | {2}        | 2    | 2
8  | 2016-06-30 | 2        | Tom Brown | {2,4}      | 3    | 3
9  | 2016-06-30 | 1        | Tom Brown | {2,4,2}    | 2    | 2
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1 回答 1

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Postgres 没有用于MEDIAN. 但是，您可以使用Postgres wiki中提供的函数片段创建一个。该片段也是ulib_agg 用户定义库的一部分。

创建后，您可以像使用任何聚合函数一样使用它，SUM或STRING_AGG具有类似window规范。windowPostgres 为您提供了为以逗号分隔的聚合函数指定多个定义的选项。

因此，要获取MEDIAN前 20 条记录中的一条，您的窗口可以在此查询中定义。

SELECT 
j.* ,  array_agg(position) over w as previous_positions,
       median(position)    over w_20 as med_20
  FROM jockeys j
WINDOW w as
(  partition by name ORDER BY id rows between 
     unbounded preceding and 1 preceding
     ),
     w_20 as
     (  partition by name ORDER BY id rows between 
           20 preceding and 1 preceding
     )

ROUND最重要的是，如果您想截断十进制数字，您可以应用函数。

演示

于 2019-05-07T15:25:11.167 回答