联合查询在一列上不同

Question

我希望第二个查询的结果覆盖第一个查询的结果：

SELECT "panel_restaurants_restaurant"."id",
       "panel_restaurants_restaurant"."name",
       "panel_restaurants_restaurant"."logo",
       "panel_restaurants_restaurantfeatures"."currency" AS "currency",
       ST_DistanceSphere(location, ST_GeomFromText('POINT(0.0 0.0)',4326)) AS "distance",
       "panel_meals_meal"."id" AS "meal_id",
       "panel_meals_meal"."status" AS "meal_status",
       "panel_meals_meal"."available_count" AS "available_dishes",
       "panel_meals_meal"."discount_price" AS "discount_price",
       "panel_meals_meal"."normal_price" AS "normal_price",
       "panel_meals_meal"."collection_from" AS "pickup_from",
       "panel_meals_meal"."collection_to" AS "pickup_to",
       "panel_meals_meal"."description" AS "meal_description"
FROM "panel_restaurants_restaurant"
INNER JOIN "panel_restaurants_restaurantfeatures" ON (
    "panel_restaurants_restaurantfeatures"."restaurant_id" = "panel_restaurants_restaurant"."id")
LEFT OUTER JOIN "panel_meals_meal" ON ("panel_restaurants_restaurant"."id" = "panel_meals_meal"."restaurant_id"
                AND "panel_meals_meal"."status" = 0
                AND (
                ("panel_meals_meal"."collection_from" AT TIME ZONE 'Europe/Warsaw')::date = DATE 'today' OR
                ("panel_meals_meal"."collection_from" AT TIME ZONE 'Europe/Warsaw')::date = DATE 'tomorrow'
                )
                AND "panel_meals_meal"."collection_to" > '2017-07-29 19:33:47.992075+00:00'
                AND "panel_meals_meal"."available_count" > 0)
WHERE "panel_restaurants_restaurant"."status" = 2
UNION
SELECT "panel_restaurants_restaurant"."id",
       "panel_restaurants_restaurant"."name",
       "panel_restaurants_restaurant"."logo",
       "panel_restaurants_restaurantfeatures"."currency" AS "currency",
       ST_DistanceSphere(location, ST_GeomFromText('POINT(0.0 0.0)',4326)) AS "distance",
       "panel_meals_meal"."id" AS "meal_id",
       "panel_meals_meal"."status" AS "meal_status",
       "panel_meals_meal"."initial_count" AS "available_dishes",
       "panel_meals_meal"."discount_price" AS "discount_price",
       "panel_meals_meal"."normal_price" AS "normal_price",
       "panel_meals_meal"."collection_from" AS "pickup_from",
       "panel_meals_meal"."collection_to" AS "pickup_to",
       "panel_meals_meal"."description" AS "meal_description"
FROM "panel_restaurants_restaurant"
INNER JOIN "panel_restaurants_restaurantfeatures" ON (
       "panel_restaurants_restaurantfeatures"."restaurant_id" = "panel_restaurants_restaurant"."id")
LEFT OUTER JOIN "panel_meals_meal" ON (
    "panel_restaurants_restaurant"."id" = "panel_meals_meal"."restaurant_id" AND
    "panel_meals_meal"."status" = 0)
INNER JOIN "panel_meals_mealrepeater" ON (
    "panel_meals_mealrepeater"."meal_id" = "panel_meals_meal"."id")
WHERE "panel_restaurants_restaurant"."status" = 2    AND "panel_meals_mealrepeater"."saturday" = true
ORDER BY distance ASC

例如 - 第一个查询可能 return 来自 panel_meals_meal table 的空值，但第二个查询会 return 某些东西 - 在那种情况下我id、name、logo、currency、distance 和不同的值（空 return 来自第一个查询, 和 something 来自另一个）对于所有其他的。

所以问题是 - 我如何使这个 UNION 在一定范围的列上不同（实际上只有一个就足够了 - id）？

Answer 1

您可以通过 FULL OUTER JOIN 而不是 UNION 来做您想做的事，并使用 COALESCE 来发挥您的优势。

我简化了您的场景以专注于 FULL OUTER JOIN 部分：

这是表格（将它们视为 第一个 SELECT 的结果 UNION，以及UNION之后的秒SELECT）：

CREATE TABLE table_a
(
    id INTEGER NOT NULL PRIMARY KEY,
    name TEXT,
    logo TEXT
) ;
CREATE TABLE table_b
(
    id INTEGER NOT NULL PRIMARY KEY,
    name TEXT,
    logo TEXT
) ;

这些是我们在其中的数据：

INSERT INTO 
   table_a
   (id, name, logo)
VALUES
   (1, 'Name1-A', 'Logo1-A'),
   (2, NULL, NULL),
   (3, 'Name3-A', NULL),
   (4, NULL, 'Logo4-A'),
   (5, 'Name5-only-in-A', NULL);

INSERT INTO 
   table_b
   (id, name, logo)
VALUES
   (1, 'Name1-B', 'Logo1-B'),
   (2, 'Name2-B', NULL),
   (3, 'Name3-B', 'Logo3-B'),
   (4, 'Name4-B', 'Logo4-B'),
   (6, 'Name6-only-in-B', 'Logo6-B');

您要查找的查询是通过加入完成的，您可以从 table_a 和 table_b 中检索所有行。然后，您使用：

SELECT
    id, 
    COALESCE(a.name, b.name) AS name,
    COALESCE(a.logo, b.logo) AS logo
FROM
    table_a AS a
    FULL OUTER JOIN table_b AS b USING(id) 
ORDER BY
    id ;

id | name            | logo   
-: | :-------------- | :------
 1 | Name1-A         | Logo1-A
 2 | Name2-B         | null   
 3 | Name3-A         | Logo3-B
 4 | Name4-B         | Logo4-A
 5 | Name5-only-in-A | null   
 6 | Name6-only-in-B | Logo6-B

dbfiddle here

在您的情况下，将 table_a AS a 替换为完整的第一个 (SELECT ...) AS a，b 也是如此。我假设 id 是您的主键。

参考文献：

Answer 2

我用 WITH query/CTE:

做了类似的事情

WITH override_query AS (SELECT * FROM blah_blah JOIN blah_blah [etc]),
     first_query AS (SELECT * FROM blah_blah JOIN blah_bluh [etc]
                    WHERE id NOT IN (SELECT id FROM override_query))
TABLE first_query UNION TABLE override_query

Answer 3

使用DISTINCT ON，例如

SELECT DISTINCT ON (maintenance_task_id)
  maintenance_task_id,
  execution_count
FROM (
  SELECT
    id maintenance_task_id,
    0 execution_count
  FROM maintenance_task
  UNION
  SELECT
    mte1.maintenance_task_id,
    count(*) execution_count
  FROM maintenance_task_execution mte1
  WHERE
    mte1.ended_at IS NULL
  GROUP BY mte1.maintenance_task_id
) AS t
ORDER BY
  maintenance_task_id,
  execution_count DESC

在此查询中：

UNION 结合了两个查询的结果。
DISTINCT ON 为每个唯一的 maintenance_task_id 值从顶部选择一行（基于 ORDER BY）。

联合查询在一列上不同

Union query distinct on one column

sql

postgresql

join

union

coalesce