Mysql查询重构案例答案

【问题标题】：Mysql Query Refactor CaseMysql查询重构案例
【发布时间】：2022-01-11 10:31:41
【问题描述】：

这是我的表结构。（MariaDB 或 MySql）

id	data
1	{"one":"1","two":"3"}
2	{"one":"2","two":"4"}

我想得到这样的输出

id	One	Two
1	Good	Bad
2	More Good	Very Bad

Values
1 = Good
2 = More Good
3 = Bad
4 = Very Bad

这是我在 MySQL 中的错误查询

SET @ONE = 'Good';
SET @TWO = 'More Good';
SET @THREE = 'Bad';
SET @FOUR = 'Very Bad';

SELECT id,

CASE
    WHEN REPLACE(json_extract(data, '$.one'), '"', '') = 1 THEN @ONE
    WHEN REPLACE(json_extract(data, '$.one'), '"', '') = 2 THEN @TWO
    WHEN REPLACE(json_extract(data, '$.one'), '"', '') = 3 THEN @THREE
    WHEN REPLACE(json_extract(data, '$.one'), '"', '') = 4 THEN @FOUR
    ELSE 'NO'
END as One,

CASE
    WHEN REPLACE(json_extract(data, '$.two'), '"', '') = 1 THEN @ONE
    WHEN REPLACE(json_extract(data, '$.two'), '"', '') = 2 THEN @TWO
    WHEN REPLACE(json_extract(data, '$.two'), '"', '') = 3 THEN @THREE
    WHEN REPLACE(json_extract(data, '$.two'), '"', '') = 4 THEN @FOUR
    ELSE 'NO'
END as Two
From TableName;

【问题讨论】：

MySQL 还是 mariaDB？现在有很多不同了！
什么是精确 DBMS版本？

标签： mysql mariadb refactoring

【解决方案1】：

您可以使用字符串函数，例如 CONCAT_WS() 来创建所有字符串变量的逗号分隔列表，并使用 SUBSTRING_INDEX() 来选择正确的值：

SET @ONE = 'Good';
SET @TWO = 'More Good';
SET @THREE = 'Bad';
SET @FOUR = 'Very Bad';

SELECT id,
       SUBSTRING_INDEX(SUBSTRING_INDEX(
         CONCAT_WS(',', @ONE, @TWO, @THREE, @FOUR),
         ',',
         json_extract(data, '$.one')
       ), ',', -1) One,
       SUBSTRING_INDEX(SUBSTRING_INDEX(
         CONCAT_WS(',', @ONE, @TWO, @THREE, @FOUR),
         ',',
         json_extract(data, '$.two')
       ), ',', -1) Two
FROM tablename;

请参阅demo。

【讨论】：

【解决方案2】：

WITH
dictionary AS ( SELECT 1 id, 'Good' val  UNION ALL
                SELECT 2,    'More Good' UNION ALL
                SELECT 3,    'Bad'       UNION ALL
                SELECT 4,    'Very Bad'  ) 
SELECT test.id, dict_1.val One, dict_2.val Two
FROM test
JOIN dictionary dict_1 ON JSON_EXTRACT(test.data, '$.one') + 0 = dict_1.id
JOIN dictionary dict_2 ON JSON_EXTRACT(test.data, '$.two') + 0 = dict_2.id

https://dbfiddle.uk/?rdbms=mariadb_10.3&fiddle=4e7cee70ca62f4c19936bc2896d1791e

PS。将“我想得到这样的输出”保存到单独的字典表中。

【讨论】：