Question

好的，我现在一直试图解决这个问题大约2个小时......请告知：

表：

PROFILE [id (int), name (varchar), ...]
SKILL   [id (int), id_profile (int), id_app (int), lvl (int), ...]
APP     [id (int), ...]

lvl基本上可以从0到3。

我正试图获得这个特定的统计数据： “至少有两个技能为2或更高的人所覆盖的应用程序的百分比是多少？”

非常感谢

Answer 1

SELECT AVG(covered)
FROM (
  SELECT CASE WHEN COUNT(*) >= 2 THEN 1 ELSE 0 END AS covered
  FROM app a
  LEFT JOIN skill s ON (s.id_app = a.id AND s.lvl >= 2)
  GROUP BY a.id
)

MySQL的更有效方式：

SELECT AVG
       (
         IFNULL
         (
           (
           SELECT 1
           FROM skill s
           WHERE s.id_app = a.id
           AND s.lvl >= 2
           LIMIT 1, 1
           ), 0
         )
       )
FROM app a

一旦找到每个person的第二位熟练app，就会停止计算。

如果您有app个person但有很多{{1}}，则效率很高。

Answer 2

未测试

select convert(float,count(*)) / (select count(*) from app) as percentage
from (
    select count(*) as number
    from skill
    where lvl >= 2
    group by id_app ) t
where t.number >= 2

Answer 3

逻辑是：百分比= 100 *（感兴趣的应用数量）/（应用总数）

select 'percentage' = 
-- 100 times
  ( cast( 100 as float ) * 
-- number of apps of interest
  ( select count(id_app) 
    from ( select id_app, count(*) as skilled_count
           from skill
           where lvl >= 2
           group by id_app 
           having count(*) >= 2 ) app_counts ) 
-- divided by total number of apps
  / ( select count(*) from app )

需要转换为float，因此sql不只是执行整数运算。

Answer 4

SELECT SUM( CASE lvl WHEN 3 THEN 1 WHEN 2 THEN 1 ELSE 0 END ) / SUM(1) FROM SKILL

如果您的数据库具有if / then函数而不是CASE，请使用它。例如，在MySQL中：

SELECT SUM( IF( lvl >= 2, 1, 0 ) ) / SUM(1) FROM SKILL

Answer 5

我不确定这是否比tvanfosson的答案更好或更差，但无论如何它仍然存在：

SELECT convert(float, count(*)) / (Select COUNT(id) FROM APP) AS percentage
FROM APP INNER JOIN SKILL ON APP.id = SKILL.id 
WHERE (
   SELECT COUNT(id) 
   FROM SKILL AS Skill2 WHERE Skill2.id_app = APP.id and lvl >= 2
) >= 2

使用复杂的SQL查询覆盖百分比...？

5 个答案: