Question

我在Postgres中的updates表是9.4.5，如下所示：

goal_id    | created_at | status
1          | 2016-01-01 | green
1          | 2016-01-02 | red
2          | 2016-01-02 | amber

这样的goals表：

id | company_id
1  | 1
2  | 2

我想为每家公司创建一个图表，每周显示所有目标的状态。

我认为这需要生成一系列过去8周，找到该周之前每个目标的最新更新，然后计算找到的更新的不同状态。

到目前为止我所拥有的：

SELECT EXTRACT(year from generate_series) AS year, 
       EXTRACT(week from generate_series) AS week,
       u.company_id,
       COUNT(*) FILTER (WHERE u.status = 'green') AS green_count,
       COUNT(*) FILTER (WHERE u.status = 'amber') AS amber_count,
       COUNT(*) FILTER (WHERE u.status = 'red') AS red_count
FROM generate_series(NOW() - INTERVAL '2 MONTHS', NOW(), '1 week')
LEFT OUTER JOIN (
  SELECT DISTINCT ON(year, week)
         goals.company_id,
         updates.status, 
         EXTRACT(week from updates.created_at) week,
         EXTRACT(year from updates.created_at) AS year,
         updates.created_at 
  FROM updates
  JOIN goals ON goals.id = updates.goal_id
  ORDER BY year, week, updates.created_at DESC
) u ON u.week = week AND u.year = year
GROUP BY 1,2,3

但这有两个问题。似乎u上的联接没有像我想象的那样起作用。它似乎是从内部查询返回的每一行（？）加入，并且这只选择从该周发生的最新更新。如果需要，它应从该周之前获取最新更新。

这是一些相当复杂的SQL，我喜欢关于如何将它拉下来的一些输入。

表格结构和信息

目标表约有1000个ATM目标，并且每周增加约100个：

                                           Table "goals"
     Column      |            Type             |                         Modifiers
-----------------+-----------------------------+-----------------------------------------------------------
 id              | integer                     | not null default nextval('goals_id_seq'::regclass)
 company_id      | integer                     | not null
 name            | text                        | not null
 created_at      | timestamp without time zone | not null default timezone('utc'::text, now())
 updated_at      | timestamp without time zone | not null default timezone('utc'::text, now())
Indexes:
    "goals_pkey" PRIMARY KEY, btree (id)
    "entity_goals_company_id_fkey" btree (company_id)
Foreign-key constraints:
    "goals_company_id_fkey" FOREIGN KEY (company_id) REFERENCES companies(id) ON DELETE RESTRICT

updates表约有1000左右，每周约有100个增长：

                                         Table "updates"
   Column   |            Type             |                            Modifiers
------------+-----------------------------+------------------------------------------------------------------
 id         | integer                     | not null default nextval('updates_id_seq'::regclass)
 status     | entity.goalstatus           | not null
 goal_id    | integer                     | not null
 created_at | timestamp without time zone | not null default timezone('utc'::text, now())
 updated_at | timestamp without time zone | not null default timezone('utc'::text, now())
Indexes:
    "goal_updates_pkey" PRIMARY KEY, btree (id)
    "entity_goal_updates_goal_id_fkey" btree (goal_id)
Foreign-key constraints:
    "updates_goal_id_fkey" FOREIGN KEY (goal_id) REFERENCES goals(id) ON DELETE CASCADE

 Schema |       Name        | Internal name | Size | Elements | Access privileges | Description
--------+-------------------+---------------+------+----------+-------------------+-------------
 entity | entity.goalstatus | goalstatus    | 4    | green   +|                   |
        |                   |               |      | amber   +|                   |
        |                   |               |      | red      |                   |

Answer 1

您每周需要一个数据项目和目标（在汇总每家公司的计数之前）。这是CROSS JOIN和generate_series()之间的简单goals。（可能）昂贵的部分是从每个state获取当前updates。与@Paul already suggested一样，LATERAL加入似乎是最好的工具。但是，仅对updates执行此操作，并使用LIMIT 1更快的技术。

使用date_trunc()简化日期处理。

SELECT w_start
     , g.company_id
     , count(*) FILTER (WHERE u.status = 'green') AS green_count
     , count(*) FILTER (WHERE u.status = 'amber') AS amber_count
     , count(*) FILTER (WHERE u.status = 'red')   AS red_count
FROM   generate_series(date_trunc('week', NOW() - interval '2 months')
                     , date_trunc('week', NOW())
                     , interval '1 week') w_start
CROSS  JOIN goals g
LEFT   JOIN LATERAL (
   SELECT status
   FROM   updates
   WHERE  goal_id = g.id
   AND    created_at < w_start
   ORDER  BY created_at DESC
   LIMIT  1
   ) u ON true
GROUP  BY w_start, g.company_id
ORDER  BY w_start, g.company_id;

要使快速，您需要多列索引：

CREATE INDEX updates_special_idx ON updates (goal_id, created_at DESC, status);

created_at的降序最好，但并非绝对必要。 Postgres几乎可以快速地向后扫描索引。（Not applicable for inverted sort order of multiple columns, though.）

订单中的索引列。为什么呢？

Multicolumn index and performance

仅添加第三列status以允许updates上的快速index-only scans。相关案例：

Slow index scans in large table

9个星期的1k目标（2个月的间隔与至少9周重叠）仅需要9k索引查找仅第1行的第2个表。对于像这样的小表，性能应该不是很大的问题。但是，如果每张表中还有几千个，则顺序扫描会降低性能。

w_start代表每周的开始。因此，计数是在一周的开始。如果你坚持，你可以仍然提取年份和周（或任何其他细节代表你的一周）：

   EXTRACT(isoyear from w_start) AS year
 , EXTRACT(week    from w_start) AS week

最好用ISOYEAR，就像@Paul解释的那样。

SQL Fiddle.

相关：

Answer 2

这似乎是LATERAL联接的好用：

SELECT  EXTRACT(ISOYEAR FROM s) AS year,
        EXTRACT(WEEK FROM s) AS week,
        u.company_id,
        COUNT(u.goal_id) FILTER (WHERE u.status = 'green') AS green_count,
        COUNT(u.goal_id) FILTER (WHERE u.status = 'amber') AS amber_count,
        COUNT(u.goal_id) FILTER (WHERE u.status = 'red') AS red_count
FROM    generate_series(NOW() - INTERVAL '2 months', NOW(), '1 week') s(w)
LEFT OUTER JOIN LATERAL (
  SELECT  DISTINCT ON (g.company_id, u2.goal_id) g.company_id, u2.goal_id, u2.status
  FROM    updates u2
  INNER JOIN goals g
  ON      g.id = u2.goal_id
  WHERE   u2.created_at <= s.w
  ORDER BY g.company_id, u2.goal_id, u2.created_at DESC
) u 
ON true
WHERE   u.company_id IS NOT NULL
GROUP BY year, week, u.company_id
ORDER BY u.company_id, year, week
;

顺便说一句，我正在提取ISOYEAR而非YEAR，以确保我在1月初获得明智的结果。例如，EXTRACT(YEAR FROM '2016-01-01 08:49:56.734556-08')为2016，EXTRACT(WEEK FROM '2016-01-01 08:49:56.734556-08')为53！

编辑：您应该测试一下您的实际数据，但我觉得这应该更快：

SELECT  year,
        week,
        company_id,
        COUNT(goal_id) FILTER (WHERE last_status = 'green') AS green_count,
        COUNT(goal_id) FILTER (WHERE last_status = 'amber') AS amber_count,
        COUNT(goal_id) FILTER (WHERE last_status = 'red') AS red_count
FROM    (
  SELECT  EXTRACT(ISOYEAR FROM s) AS year,
          EXTRACT(WEEK FROM s) AS week,
          u.company_id,
          u.goal_id,
          (array_agg(u.status ORDER BY u.created_at DESC))[1] AS last_status
  FROM    generate_series(NOW() - INTERVAL '2 months', NOW(), '1 week') s(t)
  LEFT OUTER JOIN ( 
    SELECT  g.company_id, u2.goal_id, u2.created_at, u2.status
    FROM    updates u2
    INNER JOIN goals g 
    ON      g.id = u2.goal_id
  ) u 
  ON      s.t >= u.created_at
  WHERE   u.company_id IS NOT NULL
  GROUP BY year, week, u.company_id, u.goal_id
) x
GROUP BY year, week, company_id
ORDER BY company_id, year, week
;

但仍然没有窗口功能。 :-)另外，通过用(array_agg(...))[1]函数替换first，你可以加快速度。您必须自己定义，但Postgres维基上的实现很容易谷歌。

Answer 3

我使用PostgreSQL 9.3。我对你的问题很感兴趣。我检查了你的数据结构。比我创建下表。

我插入以下记录;

公司

目标

更新

之后我编写了以下查询，进行更正

SELECT c.id company_id, c.name company_name, u.status goal_status, 
         EXTRACT(week from u.created_at) goal_status_week,
         EXTRACT(year from u.created_at) AS goal_status_year 
FROM company c
INNER JOIN goals g ON g.company_id = c.id 
INNER JOIN updates u ON u.goal_id = g.id
ORDER BY goal_status_year DESC, goal_status_week DESC;

我得到以下结果;

最后，我将此查询与周系列

合并

SELECT
             gs.company_id,
             gs.company_name,
             gs.goal_status,
             EXTRACT(year from w) AS year, 
       EXTRACT(week from w) AS week,
             COUNT(gs.*) cnt
FROM generate_series(NOW() - INTERVAL '3 MONTHS', NOW(), '1 week') w
LEFT JOIN(
SELECT c.id company_id, c.name company_name, u.status goal_status, 
             EXTRACT(week from u.created_at) goal_status_week,
       EXTRACT(year from u.created_at) AS goal_status_year 
FROM company c
INNER JOIN goals g ON g.company_id = c.id 
INNER JOIN updates u ON u.goal_id = g.id ) gs 
ON gs.goal_status_week = EXTRACT(week from w) AND gs.goal_status_year = EXTRACT(year from w)
GROUP BY company_id, company_name, goal_status, year, week
ORDER BY  year DESC, week DESC;

我得到了这个结果

祝你有个美好的一天。

每周汇总最近加入的记录

表格结构和信息

3 个答案: