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-第一查询可能返回空值什么来自于panel_meals_meal表，但第二个将返回的东西-在这种情况下我都会有相同的值id，name，logo，currency，distance和不同的值（从第一个查询返回空值，和something从另一个）。

因此，问题是-如何UNION在一定范围的列中做到这一点（实际上只有一个就足够了- id）？

Answer 1

您可以使用FULL OUTER JOIN而不是来做您想做的事情UNION，并利用COALESCE自己的优势。

我简化您的方案以专注于以下方面FULL OUTER JOIN：

这是表（想想它们是您在whe之前的第一个表，而在after之后的第二个表的结果）： SELECTUNION SELECTUNION

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');

要查找的查询是通过联接完成的，您可以从table_a和中检索所有行table_b。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 | 名称| 商标   
-：| ：-------------- | ：------
 1 | 名称1-A | 徽标1-A
 2 | 名称2-B | 空值   
 3 | 名称3-A | 徽标3-B
 4 | 名称4-B | 徽标4-A
 5 | Name5-in-A | 空值   
 6 | Name6-only-in-B | 徽标6-B

dbfiddle 在这里

在您的情况下，请以table_a AS a全额第一名代替(SELECT ...) AS a，并以代替b。我假设id是您的主键。

参考文献：

• FULL OUTER JOIN
• COALESCE
• FROM子句（查找USING）