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 - 第一个查询可能会返回来自panel_meals_meal表的内容的空值，但第二个查询会返回一些内容 - 在这种情况下，我将为id name提供相同的值1}}，logo，currency，distance和不同的值（从第一个查询返回的空值，以及来自另一个查询的something）用于所有其他值。

所以问题是 - 如何在某个列范围内使UNION明显不同（实际上只有一个就足够了 - id）？

Answer 1

您可以通过FULL OUTER JOIN代替UNION来执行您想要的操作，并使用COALESCE为您带来优势。

我简化了您的方案，专注于FULL OUTER JOIN部分：

这是表格（在SELECT之前将其视为第一个 UNION的结果，以及秒 SELECT之后说UNION）：

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，这样您就可以检索table_a和table_b中的所有行。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，请先用table_a AS a替换(SELECT ...) AS a，b替换id。我认为USING是您的主要关键。

参考文献：

FULL OUTER JOIN

COALESCE

The FROM Clause（寻找<script> $('#save_appointment').click(function () { addAppointmentInternal(); }); function addAppointmentInternal() { $.ajax({ type: 'Post', dataType: 'Json', data: { Start: $('#startAppointment').val(), End: $('#endAppointment').val(), Title: $('#title').val() }, url: '@Url.Action("AddingInternalAppointment","Calendar")', sucess: function (da) { if (da.Result === "Success") { alert(); } else { alert('Error' + da.Message); } }, error: function(da) { alert('Error'); } }); }）

Answer 2

我通过使用WITH query/CTE做过类似的事情：

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

联合查询在一列上不同

3 个答案: