MariaDB JSON_ARRAYAGG 给出错误的结果 我在 MariaDB 15.1 中使用 JSON_ARRAYAGG 时有两个问题我做错了什么？

如何解决MariaDB JSON_ARRAYAGG 给出错误的结果 我在 MariaDB 15.1 中使用 JSON_ARRAYAGG 时有两个问题我做错了什么？

我在 MariaDB 15.1 中使用 JSON_ARRAYAGG 时有两个问题

---

省略了括号 []

不正确的错误结果，值重复或省略

---

我的数据库如下：

用户：

+----+------+
| id | name |
+----+------+
|  1 | Jhon |
|  2 | Bob  |
+----+------+

汽车：

+----+---------+-------------+
| id | user_id | model       |
+----+---------+-------------+
|  1 |       1 | Tesla       |
|  2 |       1 | Ferrari     |
|  3 |       2 | Lamborghini |
+----+---------+-------------+

电话：

+----+---------+----------+--------+
| id | user_id | company  | number |
+----+---------+----------+--------+
|  1 |       1 | Verzion  |      1 |
|  2 |       1 | AT&T     |      2 |
|  3 |       1 | T-Mobile |      3 |
|  4 |       2 | Sprint   |      4 |
|  5 |       1 | Sprint   |      2 |
+----+---------+----------+--------+

---

1。省略括号 []

例如这个查询获取用户的汽车列表：

SELECT
  user.id   AS id,user.name AS name,JSON_ARRAYAGG(
    JSON_OBJECT(
      'id',car.id,'model',car.model
    )
  ) AS cars
FROM user
INNER JOIN car ON user.id = car.user_id
GROUP BY user.id;

结果：[] 中省略了方括号 cars（JSON_ARRAYAGG 的行为类似于 GROUP_CONCAT）

+----+------+-----------------------------------------------------------+
| id | name | cars                                                      |
+----+------+-----------------------------------------------------------+
|  1 | Jhon | {"id": 1,"model": "Tesla"},{"id": 2,"model": "Ferrari"} |
|  2 | Bob  | {"id": 3,"model": "Lamborghini"}                         |
+----+------+-----------------------------------------------------------+

但是当添加过滤器 WHERE user.id = 1 时，括号 [] 不会被省略：

+----+------+-------------------------------------------------------------+
| id | name | cars                                                        |
+----+------+-------------------------------------------------------------+
|  1 | Jhon | [{"id": 1,"model": "Ferrari"}] |
+----+------+-------------------------------------------------------------+

---

2.不正确的错误结果，值重复或省略

这个错误很奇怪，因为必须满足以下条件：

• 查询超过 2 个表
• 必须使用 disTINCT 选项
• 用户至少拥有 2 辆车和至少 3 部手机。

重复值

例如，此查询可获取用户的汽车列表和电话列表：

SELECT
  user.id   AS id,JSON_ARRAYAGG( disTINCT
    JSON_OBJECT(
      'id',car.model
    )
  ) AS cars,JSON_ARRAYAGG( disTINCT
    JSON_OBJECT(
        'id',phone.id,'company',phone.company,'number',phone.number
    )
  ) AS phones
FROM  user
INNER JOIN car   ON user.id =   car.user_id
INNER JOIN phone ON user.id = phone.user_id
GROUP BY user.id;

我会将输出保留为 json 格式，只保留感兴趣的元素。

结果：括号[]被省略并重复Verizon

{
  "id": 1,"name": "Jhon","phones": // [ opening bracket expected
    {
      "id": 5,"company": "Sprint","number": 2
    },{
      "id": 1,"company": "Verzion","number": 1
    },// Duplicate object with the disTINCT option
    {
      "id": 2,"company": "AT&T",{
      "id": 3,"company": "T-Mobile","number": 3
    }
  // ] Closing bracket expected
}

省略的值

在查询中省略 omit phone.id 时发生此错误

SELECT
  user.id   AS id,JSON_ARRAYAGG( disTINCT
    JSON_OBJECT(
        --'id',phone.number
    )
  ) AS phones
FROM  user
INNER JOIN car   ON user.id =   car.user_id
INNER JOIN phone ON user.id = phone.user_id
GROUP BY user.id;

结果：括号[]被省略，Sprint被省略。

显然这是因为它在 JSON_OBJECT 的列之间创建了 OR 类型，因为 company 存在于不同的行中，而 number 存在于另一个不同的行中

{
  "id": 1,"phones": // [ opening bracket expected
  //{ 
  //  "company": "Sprint",//  "number": 2
  //},`Sprint` was omitted
    {
      "company": "Verzion",{
      "company": "AT&T",{
      "company": "T-Mobile","number": 3
    }
  // ] Closing bracket expected
}

GROUP_CONCAT JSON_ARRAYAGG 实例解决了 duplicate 或 omitted 对象的问题

但是，通过添加过滤器WHERE user.id = 1，方括号[]不会被省略，也解决了duplicate或omitted对象的问题：

{
  "id": 1,"phones": [
    {
      "id": 1,{
      "id": 2,"number": 3
    },{
      "id": 5,"number": 2
    }
  ]
}

---

我做错了什么？

解决方法

到目前为止，我的解决方案是这样，但我想使用 JSON_ARRAYAGG，因为查询更清晰

-- 1
SELECT
  user.id   AS id,user.name AS name,CONCAT(
    '[',GROUP_CONCAT( DISTINCT   
      JSON_OBJECT(
        'id',car.id,'model',car.model
      )
    ),']'
  ) AS cars
FROM  user
INNER JOIN car   ON user.id =   car.user_id
GROUP BY user.id;

-- 2

SELECT
  user.id   AS id,']'
  ) AS cars,GROUP_CONCAT( DISTINCT 
      JSON_OBJECT(
          'id',phone.id,'company',phone.company,'number',phone.number
      )
    ),']'
  ) AS phones
FROM  user
INNER JOIN car   ON user.id =   car.user_id
INNER JOIN phone ON user.id = phone.user_id
GROUP BY user.id;