Question

我需要从MySQL的JSON数组中删除多个值

我尝试对每个值使用JSON_SEARCH进行JSON_REMOVE，但是在删除元素后INDEX更改

JSON数组：

<ParentFunction>g__LocalFunction|1()

例如：删除“ 5”和“ 68”

预期结果：

["1", "5", "18", "68"]

已编辑：

我尝试过的代码：

["1", "18"]

查询结果：

JSON_REMOVE(`can_see`, JSON_UNQUOTE(JSON_SEARCH(`can_see`, 'one', '5')), JSON_UNQUOTE(JSON_SEARCH(`can_see`, 'one', '68')))

在索引$ [0]处删除“ 5”后，“ 68”索引更改为$ [2]，但JSON_SEARCH从原始json返回$ [3]

我也尝试过嵌套：

["1", "18", "68"]

可以，但是如果我想删除两个以上的值，则会变得混乱

Answer 1

我需要确保数组已排序

不是最容易理解的方法，但是我认为您将必须使用SQL数字生成器将json数组解析为令牌（记录），您可以更轻松地对其进行过滤和排序。

查询

SELECT 
  JSON_ARRAYAGG(
     JSON_EXTRACT(records.json, CONCAT('$[', number_generator.number , ']'))
  ) AS json                                    
FROM (

  SELECT 
   @row := @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 @row := -1 
  ) init_user_params 
) AS number_generator
CROSS JOIN (


SELECT 
    json
  , JSON_LENGTH(records.json) AS json_array_length    
FROM (

  SELECT 
   '["1", "5", "18", "68"]' AS json
  FROM 
   DUAL  
) AS records

) AS records 
WHERE
    number BETWEEN 0 AND  json_array_length - 1   
  AND
    JSON_EXTRACT(records.json, CONCAT('$[', number_generator.number , ']')) NOT IN(5, 68)                   
ORDER BY 
 REPLACE(JSON_EXTRACT(records.json, CONCAT('$[', number_generator.number , ']')), '"', '')

结果

| json        |
| ----------- |
| ["1", "18"] |

请参阅demo

另一方面，MySQL 8+使它变得更加简单

查询

SELECT 
 JSON_ARRAYAGG (
    result_table.item
 ) AS json
FROM JSON_TABLE(
     '["1", "5", "18", "68"]'
   , "$[*]"

   COLUMNS (
       rowid FOR ORDINALITY
     , item VARCHAR(100) PATH "$"   
   )
) AS result_table
WHERE
 CAST(result_table.item AS UNSIGNED) NOT IN(5, 68) 
ORDER BY 
 CAST(result_table.item AS UNSIGNED) ASC

结果

| json        |
| ----------- |
| ["1", "18"] |

请参阅demo

如何从MySQL JSON中删除多个值

1 个答案: