mysql - 计算整个集合数据库中唯一元素的数量

Question

我有以下用户数据库，每个用户都可以说不同级别的不同语言。

id      langs
12      EN-21
36      EN-2,RU-3
41      EN-9
57      DE-35,EN-28
60      DE-9,RU-14

我想创建 MySQL 查询来计算每种语言的出现次数不管其级别。 所需的选项卡应如下所示:

lang    count
EN      4
DE      2
RU      2

我已经尝试了不同的组合，但它远非完美。

SELECT 
    DISTINCT SUBSTRING_INDEX(langs, '-', 1) AS lang, 
--  COUNT(langs) as count
--  SUM(
--      (SELECT DISTINCT SUBSTRING_INDEX(langs, '-', 1) 
--      FROM people
--      WHERE langs != '')
--  )
FROM people
WHERE langs != ''
--  GROUP BY lang
ORDER BY lang

Answer 1

如果集合中的语言数量有最大限制，你可以把第一个元素、第二个元素、第三个元素等都拉出来，并把它们联合在一起。这是一个从语言集中提取任何第一个或第二个元素并将它们组合起来的示例:

select distinct substring_index(langs, '-', 1) as lang
from people where langs != ''
union
select distinct SUBSTRING_INDEX(SUBSTRING_INDEX(langs, '-', 2), ',', -1)
from people where LENGTH(langs) - LENGTH(REPLACE(langs,',','')) + 1 > 1

演示:http://www.sqlfiddle.com/#!2/b86f2/1

---

从那里开始，将语言列表与人员列表相结合并计算匹配次数，通过比较 people.langs like '%EN%' 例如:

select
  lang,
  count(case when people.langs like concat('%',langs.lang,'%') then 1 end) as count
from people,
  (
    select distinct substring_index(langs, '-', 1) as lang
    from people where langs != ''
    union
    select distinct SUBSTRING_INDEX(SUBSTRING_INDEX(langs, '-', 2), ',', -1)
    from people where LENGTH(langs) - LENGTH(REPLACE(langs,',','')) + 1 > 1
  ) langs
group by langs.lang
order by langs.lang

示例输出:

LANG    COUNT
====    ====
DE      2
EN      4
RU      2

演示:http://www.sqlfiddle.com/#!2/b86f2/5