这是 PostgreSQL 中视图的正确用法吗？

Question

我有两个 table。一个存储位置，另一个存储这些位置的不同评级（评级可以来自不同的用户，因此一个位置可以有多个评级）。我想要一种在平均评分旁边显示位置名称的方法。我相信我需要使用视图来创建它。这是我的 tables:

locations 有一个 name 属性和一个 id 属性这是我 ratings [=36= 的外键] location_id.
ratings 有一个 location_id 属性、一个 user_id 属性和保持 eating、[=20 评级的属性=]、child_friendly、pet_friendly 和 accessible.

这是我创建的视图：

CREATE VIEW v_location_averages AS
SELECT
    r.location_id,
    t.name,
    t.eating,
    t.reading,
    t.child_friendly,
    t.pet_friendly,
    t.accessible,
    (COALESCE(t.eating, 0) + COALESCE(t.reading, 0) + COALESCE(t.child_friendly, 0) + COALESCE(t.pet_friendly, 0) + COALESCE(t.accessible, 0)) / num_nonnulls(t.eating, t.reading, t.child_friendly, t.pet_friendly, t.accessible) AS "overall_rating"
FROM (
    SELECT 
        locations.name,
        location_id, 
        AVG(eating) AS "eating",
        AVG(reading) AS "reading",
        AVG(child_friendly) AS "child_friendly",
        AVG(pet_friendly) AS "pet_friendly",
        AVG(accessible) AS "accessible",
    FROM ratings 
    INNER JOIN locations on ratings.location_id = locations.id
    GROUP BY location_id, locations.name
) t
JOIN ratings r ON r.location_id = t.location_id
GROUP BY r.location_id, t.name, t.eating, t.reading, t.child_friendly, t.pet_friendly, t.accessible;

这是实现我想要的目标的正确方法吗？这是创建该视图的最佳方法吗？（确实有效，但正确吗？）

Answer 1

使用视图确实是一种很好的做法，但在您的情况下，视图可以更简单：

CREATE VIEW v_location_averages AS
SELECT
    t.location_id,
    l.name,
    t.eating,
    t.reading,
    t.child_friendly,
    t.pet_friendly,
    t.accessible,
    (COALESCE(t.eating, 0) + COALESCE(t.reading, 0) + COALESCE(t.child_friendly, 0) + COALESCE(t.pet_friendly, 0) + COALESCE(t.accessible, 0)) / num_nonnulls(t.eating, t.reading, t.child_friendly, t.pet_friendly, t.accessible) AS "overall_rating"
FROM (
    SELECT 
        location_id, 
        AVG(eating) AS "eating",
        AVG(reading) AS "reading",
        AVG(child_friendly) AS "child_friendly",
        AVG(pet_friendly) AS "pet_friendly",
        AVG(accessible) AS "accessible",
    FROM ratings 
    GROUP BY location_id
) t 
INNER JOIN locations l on t.location_id = l.id;

这是 PostgreSQL 中视图的正确用法吗？

Is this the correct usage of a view in PostgreSQL?

postgresql

database-design