如何将 3 个查询合二为一，在每个表中加入并计数一行？

Question

我的桌子..

users
---------------------
pass | name | genre
---------------------
 1   | Mike | pop
 2   | Clark| rock


favorites
---------------------
id | pass | songs
---------------------
 1 | 1    | blabla
 2 | 1    | blaablaa
 3 | 1    | blblbl
 4 | 2    | lalala


suggestions
---------------------
id | pass | songs
---------------------
 1 | 1    | brrbrrr
 2 | 2    | lulala

我的查询..

1.对于用户名和类型

SELECT `name`, `genre` FROM `users` WHERE `name` = "Mike"
$user_data = mysqli_fetch_object($query);
$user_data->name;
$user_data->genre;

2.数数“Mike”中最喜欢的歌曲

SELECT COUNT(`songs`) AS "total_favorites" FROM `favorites` INNER JOIN `users` USING (pass) WHERE `name` = "Mike"
$fav_dat = mysqli_fetch_object($query);
$fav_dat->total_favorites;

3.数数来自“Mike”的建议歌曲

SELECT COUNT(`songs`) AS "total_suggestions" FROM `suggestions` INNER JOIN `users` USING (pass) WHERE `name` = "Mike"
$sug_dat = mysqli_fetch_object($query);
$sug_dat->total_suggestions;

如何将这 3 个查询合二为一？有没有更好的写法？

Answer 1

您可以对子查询使用连接

SELECT users.name, `genre`, t1.total_favorites, t2.total_suggestions
FROM `users` 
INNER JOIN  (
    SELECT name,  COUNT(`songs`) AS total_favorites 
    FROM `favorites` 
    INNER JOIN `users` USING (pass) 
    GROUP BY name
    
) t1 ON t1.name  = users.name
INNER JOIN  (
    SELECT name,  COUNT(`songs`) AS total_suggestions 
    FROM `suggestions` 
    INNER JOIN `users` USING (pass) 
    GROUP BY name
) t2 t2.name  = users.name
WHERE users.name = "Mike"

对于不匹配的值，您可以尝试 uning left join 和 ifnull

SELECT users.name, `genre`, ifnull(t1.total_favorites,0) total_favorites, ifnull(t2.total_suggestions,0) total_suggestions
FROM `users` 
LEFT  JOIN  (
    SELECT name,  COUNT(`songs`) AS total_favorites 
    FROM `favorites` 
    INNER JOIN `users` USING (pass) 
    GROUP BY name
    
) t1 ON t1.name  = users.name
LEFT  JOIN  (
    SELECT name,  COUNT(`songs`) AS total_suggestions 
    FROM `suggestions` 
    INNER JOIN `users` USING (pass) 
    GROUP BY name
) t2 t2.name  = users.name
WHERE users.name = "Mike"