这需要使用MySQL 8.0中支持的JSON_TABLE（）函数：

mysql> select id, j.tag from mytable 
  left join json_table(tags, '$[*]' columns( tag varchar(10) path '$' )) j 
  on true; 
+------+------+
| id   | tag  |
+------+------+
|    1 | a    |
|    1 | b    |
|    2 | a    |
|    2 | b    |
|    2 | c    |
|    3 | NULL |
+------+------+

mysql> select id, group_concat(j.tag) as tags from mytable 
  left join json_table(tags, '$[*]' columns( tag varchar(10)
path '$' )) j 
  on true 
  group by id;
+------+-------+
| id   | tags  |
+------+-------+
|    1 | a,b   |
|    2 | a,b,c |
|    3 | NULL  |
+------+-------+

您可以这样做：

SELECT
  `id`,
  /* tags_joined */ (
    SELECT
      COALESCE(
        GROUP_CONCAT(`jsontable`.`item` SEPARATOR ', '), -- join array items
        '' -- fallback value for nonexistent or empty array
      )
    FROM
      `my_table` AS `jsonsource`, -- alias for WHERE
      -- implicit lateral join
      JSON_TABLE(
        `jsonsource`.`tags`,
        '$[*]' -- path to array items
        COLUMNS(`item` VARCHAR(10) PATH '$') -- array item to table column
      ) AS `jsontable`
    WHERE `jsonsource`.`id` = `my_table`.`id` -- current row only
  ) AS `tags_joined`
FROM `my_table`