Question

我正在处理此查询，但有些困难。联接对我来说很有意义，但是从images表（其中有一个名为user_image的联结表）中获取数据正在使我逃脱。

SELECT c.id, c.city, ROUND(( 3959 * acos( cos( radians(27.9861) ) * cos( radians( lat ) ) 
* cos( radians( lng ) - radians(-80.6628) ) + sin( radians(27.9861) ) * sin(radians(lat)) ) ),0) AS distance, t.id, t.name, t.price, t.duration, d.description, u.fname, i.image_path, i.image_name
FROM city c
    JOIN trip t ON c.id = t.city_id
    JOIN trip_description d ON t.id = d.trip_id
    JOIN user u ON t.user_id = u.id
    -- Need to get all images that match trip and is_main = 1
HAVING distance < 20 
ORDER BY distance 
LIMIT 0 , 45;

我的图像表看起来像这样...

user_image：user_id | trip_id | image_id
图片：id | image_name | is_active | is_main

不确定我是否应该在查询中使用另一个联接，联合，查询？真是茫然，不胜感激，将有所帮助：）

Answer 1

假设您要查找给定用户的所有与is_main = 1和is_active = 1匹配的图像名称：

SELECT c.id as city_id, c.city, u.fname, t.name, d.description, i.image_name
FROM image i
    LEFT JOIN user_image ui ON ui.image_id = i.id 
    LEFT JOIN user u ON u.id = ui.user_id
    LEFT JOIN trip t ON t.id = ui.trip_id
    LEFT JOIN trip_description d ON t.id = d.trip_id
    LEFT JOIN city c ON c.id = t.city_id
WHERE u.id = 1
    AND i.is_main = 1
    AND i.active = 1
ORDER BY c.id;

Answer 2

好吧，我想我明白了。感谢所有回应。

SELECT c.id, c.city, ROUND(( 3959 * acos( cos( radians(27.9861) ) * cos( radians( lat ) ) 
* cos( radians( lng ) - radians(-80.6628) ) + sin( radians(27.9861) ) * sin(radians(lat)) ) ),0) AS distance, t.id, t.name, t.price, t.duration, d.description, u.fname, i.path, i.name
FROM city c
    JOIN trip t ON c.id = t.city_id
    JOIN trip_description d ON t.id = d.trip_id
    JOIN user u ON t.user_id = u.id
    LEFT OUTER JOIN user_image ui ON ui.trip_id = t.id
    LEFT OUTER JOIN image i ON ui.image_id = i.id
    AND i.main=1
HAVING distance < 20 
ORDER BY distance 
LIMIT 0 , 45;

具有多个联接和联结表的MySQL查询

2 个答案: