mysql - SQL GROUP_CONCAT 拆分成不同的列

Question

我搜索了很多，但没有找到适合我的问题的解决方案。

我想做什么？

我在 MySQL 中有 2 个表:- 国家- 货币(我通过 CountryCurrency 将它们连接在一起 --> 由于多对多的关系)

有关工作示例，请参见:http://sqlfiddle.com/#!2/317d3/8/0

我想使用联接将两个表链接在一起，但我只想为每个国家/地区显示一行(有些国家/地区有多种货币，所以这是第一个问题)。

我找到了 group_concat 函数:

SELECT country.Name, country.ISOCode_2, group_concat(currency.name) AS currency
FROM country
INNER JOIN countryCurrency ON country.country_id = countryCurrency.country_id
INNER JOIN currency ON currency.currency_id = countryCurrency.currency_id
GROUP BY country.name

结果如下:

NAME            ISOCODE_2   CURRENCY

Afghanistan AF          Afghani
Åland Islands   AX          Euro
Albania         AL          Lek
Algeria         DZ          Algerian Dinar
American Samoa  AS          US Dollar,Kwanza,East Caribbean Dollar

但我现在想要的是将货币拆分为不同的列(货币 1、货币 2，...)。我已经尝试过像 MAKE_SET() 这样的函数，但这不起作用。

Answer 1

您可以使用 substring_index() 执行此操作。以下查询使用您的查询作为子查询，然后应用此逻辑:

select Name, ISOCode_2,
       substring_index(currencies, ',', 1) as Currency1,
       (case when numc >= 2 then substring_index(substring_index(currencies, ',', 2), ',', -1) end) as Currency2,
       (case when numc >= 3 then substring_index(substring_index(currencies, ',', 3), ',', -1) end)  as Currency3,
       (case when numc >= 4 then substring_index(substring_index(currencies, ',', 4), ',', -1) end)  as Currency4,
       (case when numc >= 5 then substring_index(substring_index(currencies, ',', 5), ',', -1) end)  as Currency5,
       (case when numc >= 6 then substring_index(substring_index(currencies, ',', 6), ',', -1) end)  as Currency6,
       (case when numc >= 7 then substring_index(substring_index(currencies, ',', 7), ',', -1) end)  as Currency7,
       (case when numc >= 8 then substring_index(substring_index(currencies, ',', 8), ',', -1) end)  as Currency8
from (SELECT country.Name, country.ISOCode_2, group_concat(currency.name) AS currencies,
             count(*) as numc
      FROM country
      INNER JOIN countryCurrency ON country.country_id = countryCurrency.country_id
      INNER JOIN currency ON currency.currency_id = countryCurrency.currency_id
      GROUP BY country.name
     ) t

表达式 substring_index(currencies, ',' 2) 获取货币列表直到第二个。对于 American Somoa，这将是 'US Dollar,Kwanza'。以 -1 作为参数的下一次调用采用列表的最后一个元素，即 'Kwanza'，它是 currencies .

另请注意，SQL 查询会返回一组明确定义的列。查询的列数不能可变(除非您通过 prepare 语句使用动态 SQL)。