以每周日期为周期对用户进行排名并列出所有排名第一的用户

Question

我有一个名为优惠券的 table，架构如下：

CREATE TABLE "public"."coupons" (
   "id" int4 NOT NULL,
    "suprise" bool NOT NULL DEFAULT false,
    "user_id" int4 NOT NULL,
    "start" timestamp NOT NULL,
    "win_price" numeric(8,2) NOT NULL DEFAULT 0::numeric,
    "fold" int4 NOT NULL DEFAULT 3,
    "pay" numeric(8,2) NOT NULL DEFAULT 0::numeric,
    "rate" numeric(8,2) NOT NULL DEFAULT 0::numeric,
    "win" varchar(255) NOT NULL DEFAULT 'H'::character varying COLLATE "default",
    "end" timestamp NOT NULL,
    "win_count" int4 NOT NULL DEFAULT 0,
    "match_count" int4 NOT NULL DEFAULT 0,
    "played" bool NOT NULL DEFAULT false,
    "created_at" timestamp NOT NULL,
    "updated_at" timestamp NOT NULL
)
WITH (OIDS=FALSE);

为了让用户排名超过 win_price weekly，我编写了下面的查询以在 2015 年 7 月 27 日和 2015 年 8 月 3 日之间获得前 5 名：

SELECT ROW_NUMBER() OVER(ORDER BY sum(win_price) DESC) AS rnk,
       sum(win_price) AS win_price, user_id,
       min(created_at) min_create
FROM coupons
WHERE played = true AND win = 'Y'
AND created_at BETWEEN '27-07-2015' AND '03-08-2015'
GROUP BY user_id
ORDER BY rnk ASC
LIMIT 5;

我正在寻找一个新查询，该查询每周列出给定日期段内排名第一的用户。
即：对于 01-09-2015 和 30-09-2015 之间的时间段：

rnk - win_price - user_id - min_create  
 1  - 1.52      - 1       - ...........  (first week)
 1  - 10.92     - 2       - ...........  (send week)
 1  - 11.23     - 1       - ...........  (third week and so on)

Answer 1

SELECT *
FROM  (
   SELECT date_trunc('week', created_at) AS week
        , rank() OVER (PARTITION BY date_trunc('week', created_at)
                       ORDER BY sum(win_price) DESC NULLS LAST) AS rnk
        , sum(win_price) AS win_price
        , user_id
        , min(created_at) min_create
   FROM   coupons
   WHERE  played = true
   AND    win = 'Y' AND created_at BETWEEN '27-07-2015' AND '03-08-2015'
   GROUP  BY 1, 4  -- reference to 1st & 4th column
   ) sub
WHERE  rnk = 1
ORDER  BY week;

这 returns 获胜用户每周 - 获得最大 sum(win_price).

的用户

请注意，我使用的是 rank() instead of row_number()，因为您没有为每周多个获胜者定义决胜局。

还要注意排序子句DESC NULLS LAST：这会阻止 NULL 值首先排序（如果你应该有 NULL）：

PostgreSQL sort by datetime asc, null first?

周由开始时间戳表示，您可以使用 to_char() 任意格式。

查询的关键要素：您可以在聚合函数上使用 window 函数。详情：

Postgres window function and group by exception

考虑 SELECT 查询中的事件序列：

Best way to get result count before LIMIT was applied

以每周日期为周期对用户进行排名并列出所有排名第一的用户

Ranking users with weekly date period and listing all first ranked users

sql

postgresql

aggregate-functions

greatest-n-per-group

window-functions