3

我有以下用户数据库，每个用户都可以在不同级别上讲不同的语言。

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，而不管其级别。所需的选项卡应如下所示：

lang    count
EN      4
DE      2
RU      2

我已经尝试过不同的组合，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
4

2 回答 2

2

如果集合中的语言数量有最大限制，则可以将第一个元素、第二个元素、第三个元素等全部拉出，并合并在一起。这是一个示例，它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：

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结合起来并计算匹配的数量，people.langs like '%EN%'例如比较：

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

样本输出：

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

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

于 2012-05-24T14:48:26.600 回答
0 
SELECT SUBSTRING_INDEX(langs, '-', 1) AS lang, count(1) as count_lang
FROM people
WHERE langs!=''
GROUP BY lang
ORDER BY lang

请试试这个，让我知道你得到了什么。

于 2012-05-24T14:31:06.527 回答