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sql - LEFT OUTER JOIN 查询不包含左表的结果

问题描述

我有一些桌子：

CREATE TABLE IF NOT EXISTS
        days(
        id INTEGER PRIMARY KEY AUTOINCREMENT,
        name TEXT,
        active BOOLEAN
        );

INSERT into days (name, active) VALUES ('S', True);
INSERT into days (name, active) VALUES ('M', True);
INSERT into days (name, active) VALUES ('T', True);
INSERT into days (name, active) VALUES ('W', True);
INSERT into days (name, active) VALUES ('Th', True);
INSERT into days (name, active) VALUES ('Fr', True);
INSERT into days (name, active) VALUES ('Sut', True);

CREATE TABLE IF NOT EXISTS
        user(
        id INTEGER PRIMARY KEY AUTOINCREMENT,
        mail TEXT,
        nickname TEXT,
        isadmin INTEGER
        );

INSERT into user (nickname, mail) VALUES ('overblown', 'abc@abc.com');


CREATE TABLE IF NOT EXISTS
        dish(
        id INTEGER PRIMARY KEY AUTOINCREMENT,
        name TEXT,
        price FLOAT
        );

INSERT into dish (name,price) VALUES ('Pizza', 40);
INSERT into dish (name,price) VALUES ('Soup', 20);
INSERT into dish (name,price) VALUES ('Bread', 40);
INSERT into dish (name,price) VALUES ('Chips', 42);


CREATE TABLE IF NOT EXISTS
        menu(
        id INTEGER PRIMARY KEY AUTOINCREMENT,
        dish_id INTEGER,
        day_id INTEGER
        );

INSERT into menu (dish_id, day_id) VALUES (1, 1);
INSERT into menu (dish_id, day_id) VALUES (2, 1);
INSERT into menu (dish_id, day_id) VALUES (3, 1);
INSERT into menu (dish_id, day_id) VALUES (4, 1);


CREATE TABLE IF NOT EXISTS
        request(
        id INTEGER PRIMARY KEY AUTOINCREMENT,
        menu_id INTEGER,
        user_id INTEGER,
        count FLOAT
        );

INSERT into request (menu_id, user_id, count) VALUES (1, 1, 0.5);
INSERT into request (menu_id, user_id, count) VALUES (2, 1, 0.5);

我想从request表中为某人用户获取所有记录，但还要将表中的所有记录添加到menu某天，以便结果相对于menu_id. Sqlite 不支持RIGHT OUTER JOIN，我尝试使用UNION：

select user.nickname as nick, dish.name as dish, days.name as day, request.count as size from request 
        INNER JOIN user on user.id = request.user_id 
        INNER JOIN menu on menu.id = request.menu_id
        INNER JOIN dish on dish.id = menu.dish_id
        INNER JOIN days on days.id = menu.day_id
        where user_id=1 and days.id=1
UNION
SELECT null as nick, dish.name as dish, days.name as day, 0.0 size from menu
        INNER JOIN dish on dish.id = menu.dish_id
        INNER JOIN days on days.id = menu.day_id
        where menu.day_id=1;

看起来不错，但结果没有唯一值。好的，我尝试使用LEFT OUTER JOIN（相对于操作员改变桌子的侧面）

SELECT user.nickname as nick, dish.name as dish, days.name as day, count
        FROM menu
        LEFT outer JOIN request ON request.menu_id=menu.id
        INNER JOIN user on user.id = request.user_id 
        INNER JOIN dish on dish.id = menu.dish_id
        INNER JOIN days on days.id = menu.day_id
        where menu.day_id=1 and request.user_id=1

但结果仅具有来自request表的行，而不包含来自menu. 我做错了什么？

预期结果：

('overblown', 'Pizza', 'S', 0.5)
('overblown', 'Soup', 'S', 0.5)
('overblown', 'Bread', 'S', NULL)
('overblown', 'Chips', 'S', NULL)

标签： sqlsqliteleft-join

解决方案

您需要在left join整个查询中使用：

SELECT u.nickname as nick, d.name as dish, dy.name as day, count
FROM request r LEFT JOIN
     menu m
     ON r.menu_id = m.menu_id AND
        m.day_id = 1 LEFT JOIN
     user u
     ON u.id = r.user_id LEFT JOIN
     dish d
     ON d.id = m.dish_id LEFT JOIN
     days dy
     ON dy.id = m.day_id
WHERE r.user_id = 1;

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