如何在MySQL 5.7上模拟JSON_重叠函数？_Mysql

如何在MySQL 5.7上模拟JSON_重叠函数？

mysql

如何在MySQL 5.7上模拟JSON_重叠函数？,mysql,Mysql,我有两个来自不同表的列，它们保存JSON格式的数据。两列中存储的数据都是数组。例如：用户 +----+------------------+ | id | options | +----+------------------+ | 1 | ["AB","CD","XY"] | | 2 | ["CD","GH"] | +----+------------------+ +----+-------------+ | id | options | +----

我有两个来自不同表的列，它们保存JSON格式的数据。两列中存储的数据都是数组。例如：

用户

+----+------------------+
| id | options          |
+----+------------------+
|  1 | ["AB","CD","XY"] |
|  2 | ["CD","GH"]      |
+----+------------------+

+----+-------------+
| id | options     |
+----+-------------+
| 10 | ["CD","EF"] |
| 11 | ["GH","XY"] |
| 12 | ["GH"]      |
+----+-------------+

项目

+----+------------------+
| id | options          |
+----+------------------+
|  1 | ["AB","CD","XY"] |
|  2 | ["CD","GH"]      |
+----+------------------+

+----+-------------+
| id | options     |
+----+-------------+
| 10 | ["CD","EF"] |
| 11 | ["GH","XY"] |
| 12 | ["GH"]      |
+----+-------------+

我想编写一个查询，使用

选项

列执行匹配，返回

用户

中与

项

中给定行匹配的所有行。规则是，如果两行中都存在数组中的任何值，则它们是匹配的。示例：用户1将匹配项目10（因为

CD

选项）和项目11（因为

XY

选项）；用户2将匹配项目10、11和12，因为它们都有

CD

或

GH

查看MySQL文档，我发现它确实做到了这一点。但是，我运行的是MySQL 5.7，该函数仅从8.0.17开始提供。网络上也没有太多关于这个功能的讨论

如何在查询中模拟MySQL 5.7上的

JSON\u重叠

行为

编辑：不幸的是，升级到MySQL 8不是一个选项，因为我们在生产环境中运行MariaDB，它也没有该功能

如何在MySQL 5.7上模拟JSON_重叠函数

编辑：不幸的是，升级到MySQL 8不是一个选项，因为我们在生产环境中运行MariaDB，它也没有这个功能

像草莓一样受到警告，建议升级更容易

SELECT 
 *
FROM (

SELECT
    items.id
  ,    JSON_UNQUOTE(
       JSON_EXTRACT(items.options, CONCAT('$[', number_generator.number , ']'))
     ) AS json_options
                     
                                
FROM (

  SELECT 
   @items_row := @items_row + 1 AS number
  FROM (
    SELECT 0 UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5 UNION   SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9
  ) row1
  CROSS JOIN (
    SELECT 0 UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5 UNION  SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9
  ) row2
  CROSS JOIN (
    SELECT @items_row := -1 
  ) init_user_params 
) AS number_generator
CROSS JOIN (


SELECT 
    items.id
  , items.options
  , JSON_LENGTH(items.options) AS json_array_length    
FROM 
   items

) AS items 
WHERE
    number BETWEEN 0 AND  json_array_length - 1   
                    
) AS items 
INNER JOIN (

  SELECT
    users.id
  ,    JSON_UNQUOTE(
       JSON_EXTRACT(users.options, CONCAT('$[', number_generator.number , ']'))
     ) AS json_options
                     
                                
FROM (

  SELECT 
   @users_row := @users_row + 1 AS number
  FROM (
    SELECT 0 UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5 UNION   SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9
  ) row1
  CROSS JOIN (
    SELECT 0 UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5 UNION  SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9
  ) row2
  CROSS JOIN (
    SELECT @users_row := -1 
  ) init_user_params 
) AS number_generator
CROSS JOIN (


SELECT 
    users.id
  , users.options
  , JSON_LENGTH(users.options) AS json_array_length    
FROM 
   users

) AS users 
WHERE
    number BETWEEN 0 AND  json_array_length - 1   
) AS users 
USING(json_options)

现在，这是不可能的。你还是自找的，让我们玩一玩吧。我过去发布了一些来模拟MySQL的8

JSON_TABLE（）

，为什么我要提到这一点？因为我使用此方法模拟MySQL的8个

JSON_重叠

，将模拟

JSON_TABLE（）

的两个结果集简单地连接到最终的结果集

这将生成下面的查询（请原谅格式）

查询

SELECT 
 *
FROM (

SELECT
    items.id
  ,    JSON_UNQUOTE(
       JSON_EXTRACT(items.options, CONCAT('$[', number_generator.number , ']'))
     ) AS json_options
                     
                                
FROM (

  SELECT 
   @items_row := @items_row + 1 AS number
  FROM (
    SELECT 0 UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5 UNION   SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9
  ) row1
  CROSS JOIN (
    SELECT 0 UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5 UNION  SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9
  ) row2
  CROSS JOIN (
    SELECT @items_row := -1 
  ) init_user_params 
) AS number_generator
CROSS JOIN (


SELECT 
    items.id
  , items.options
  , JSON_LENGTH(items.options) AS json_array_length    
FROM 
   items

) AS items 
WHERE
    number BETWEEN 0 AND  json_array_length - 1   
                    
) AS items 
INNER JOIN (

  SELECT
    users.id
  ,    JSON_UNQUOTE(
       JSON_EXTRACT(users.options, CONCAT('$[', number_generator.number , ']'))
     ) AS json_options
                     
                                
FROM (

  SELECT 
   @users_row := @users_row + 1 AS number
  FROM (
    SELECT 0 UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5 UNION   SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9
  ) row1
  CROSS JOIN (
    SELECT 0 UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5 UNION  SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9
  ) row2
  CROSS JOIN (
    SELECT @users_row := -1 
  ) init_user_params 
) AS number_generator
CROSS JOIN (


SELECT 
    users.id
  , users.options
  , JSON_LENGTH(users.options) AS json_array_length    
FROM 
   users

) AS users 
WHERE
    number BETWEEN 0 AND  json_array_length - 1   
) AS users 
USING(json_options)

结果

| json_options | id  | id  |
| ------------ | --- | --- |
| CD           | 10  | 2   |
| CD           | 10  | 1   |
| GH           | 11  | 2   |
| GH           | 12  | 2   |
| XY           | 11  | 1   |

请参见

simpler@Strawberry我想有人会建议这样做，所以我只是编辑了这个问题，告诉大家我们在生产中运行MariaDB。但无论如何，还是要谢谢你。我发布了一个答案，我相信它也应该适用于MariaDB~10.2+（我相信）…嘿！尽管这是一个巨大的疑问，但它却像一个魅力！谢谢“尽管这是一个巨大的疑问，”很抱歉，我从未说过这很容易@GustavoStraube，希望你能找到它的工作原理。。