SQL 组内排序在 Postgres 组内显示顶部

Question

我在 Postgres 中有一张表，其中包含如下数据：

fk | priority | date
1       1       4-1-2023
1       1       4-3-2023
1       2       4-5-2023
2       3       4-1-2023
2       2       4-2-2023
3       3       4-6-2023

我需要这样的结果：

fk | priority | date
1       1       4-3-2023
2       2       4-2-2023
3       3       4-6-2023

所以在 fk 的每个分组中，我想要最高优先级。 1为最高。如果有优先关系，那么我想要最近的日期。

我的 SQL 不是那么先进，我很难找到解决方案。提前致谢。

Answer 1

这可以通过带有子查询的窗口函数来完成（适用于多个 RDBMS）

SELECT fk, priority, mydate
FROM (
    SELECT *, /* Return all the columns we want in the final query */
        ROW_NUMBER() OVER (PARTITION BY fk ORDER BY priority, mydate DESC) AS rn
        /* 
            Generate a row number, 
            restarting the numbering for each `fk` value seen (partition by)
            Order the rows, first: the smallest number `priority` (ascending),
            then: the latest `mydate` (descending)
            Put the result in a column named `rn` to use in the outer query
        */
    FROM mytable
) x
WHERE rn = 1 /* Filter by the `rn` value, return only the first row */
;

或者使用

DISTINCT ON

子句（更具体的 postgresql）

SELECT DISTINCT ON (fk) * /* Return a distinct row for each fk value */
FROM mytable
ORDER BY fk, priority, mydate DESC 
/* 
    Order by which is the first row to return 
    It is required to include `fk` first because of the DISTINCT ON (fk), 
    but then: get the smallest `priority` (ascending),
    then: the latest `mydate` (descending).
*/
;