SQL一请求加入表group_concat

Posted 2023-02-24

【中文标题】SQL一请求加入表group_concat

【英文标题】：SQL one request joining table group_concat

【发布时间】：2013-12-08 01:55:04

【问题描述】：

我有这三个非常简单的表：

table user = 
[
    [
        'id'       => 1
        'username' => 'user1'
    ]
    [
        'id'       => 2
        'username' => 'user2'
    ]
    [
        'id'       => 3
        'username' => 'user3'
    ]
]

table cars = 
[
    [
        'id_user'  => 1
        'name'     => 'bmw'
    ]
    [
        'id_user'  => 1
        'name'     => 'audi'
    ]
    [
        'id_user'  => 1
        'name'     => 'honda'
    ]
    [
        'id_user'  => 2
        'name'     => 'dodge'
    ]
]

table bikes = 
[
    [
        'id_user'  => 1
        'name'     => 'yamaha'
    ]
    [
        'id_user'  => 1
        'name'     => 'suzuki'
    ]
    [
        'id_user'  => 3
        'name'     => 'kawasaki'
    ]
]

我想通过他的 id 加载一个用户，只用一个请求并得到这样的结果：

id = 1 的示例：

array result = 
[
    'id'         => 1
    'username'   => 'user1'
    'cars_name'  => 'bmw##audi##honda'
    'bikes_name' => 'yamaha##suzuki'
]

id = 2 的示例：

array result = 
[
    'id'         => 2
    'username'   => 'user2'
    'cars_name'  => 'dodge'
    'bikes_name' => ''
]

我已经尝试过这个请求：

'
    SELECT 
        user.id,
        user.username,
        GROUP_CONCAT(cars.name SEPARATOR "##"),
        GROUP_CONCAT(bikes.name SEPARATOR "##")
    FROM 
        user
    LEFT JOIN 
        cars ON user.id = cars.id_user
    LEFT JOIN
        bikes ON user.id = bikes.id_user
    WHERE
        user.id = $id
    LIMIT 1
'

但是这个输出（例如 id = 1）：

array result = 
[
    'id'         => 1
    'username'   => 'user1'
    'cars_name'  => 'bmw##audi#honda##bmw##audi#honda'
    'bikes_name' => 'yamaha##yamaha##yamaha##suzuki##suzuki##suzuki'
]

但是，它适用于 id = 2 和 id = 3 ：

(id = 2)

array result = 
[
    'id'         => 2
    'username'   => 'user2'
    'cars_name'  => 'dodge'
    'bikes_name' => ''
]

(id = 3)

array result = 
[
    'id'         => 1
    'username'   => 'user3'
    'cars_name'  => ''
    'bikes_name' => 'kawasaki'
]

我该如何解决这个问题？

感谢您的帮助！

【参考方案1】：

两个 LEFT JOIN 将结果相乘，为用户 1 提供 6 个结果（3 辆汽车 x 2 辆自行车）。

您可以在 GROUP_CONCAT 中使用 DISTINCT 来修复重复项。

来自docs：

GROUP_CONCAT([DISTINCT] expr [,expr ...]

         [ORDER BY unsigned_integer | col_name | expr
             [ASC | DESC] [,col_name ...]]
         [SEPARATOR str_val])

看起来像这样：

SELECT 
  user.id,
  user.username,
  GROUP_CONCAT(DISTINCT cars.name SEPARATOR "##") AS cars,
  GROUP_CONCAT(DISTINCT bikes.name SEPARATOR "##") AS bikes
FROM user 
LEFT JOIN cars ON user.id = cars.id_user
LEFT JOIN bikes ON user.id = bikes.id_user
GROUP BY user.id