Question

所以我目前有一个数据库可以存储调查问题，我现在想要运行一个特殊查询，但不知道该怎么做。 需要注意的事项：根据问题，每个问题可以有2-6个答案。

这些是我使用的表格以及一些示例数据：

 Table: answers_only         Table: questions_only
╔════════════════╦═══════════╗ ╔═════════════════╦════════════════════════════╗
║ answer_ID (PK) ║ answer    ║ ║question_ID (PK) ║ question                   ║
╠════════════════╬═══════════╣ ╠═════════════════╬════════════════════════════╣
║    65114       ║ yes       ║ ║       123       ║ Are you happy?             ║
║    614         ║ no        ║ ║       1967      ║ You think you're smart?    ║
║    23          ║ sometimes ║ ╚═════════════════╩════════════════════════════╝      
╚════════════════╩═══════════╝

                    Table: questions
╔════════════════╦══════════════════╦════════════════╦════════════════╗
║ unique_ID (PK) ║ question_ID (FK) ║ answer_ID (FK) ║ person_ID (FK) ║
╠════════════════╬══════════════════╬════════════════╬════════════════╣
║              1 ║              123 ║          65114 ║           5521 ║
║              2 ║              123 ║          614   ║           2511 ║
║              3 ║             1967 ║          614   ║           2511 ║
╚════════════════╩══════════════════╩════════════════╩════════════════╝

所以我有一个表格问题，其中包含 questions_only 的ID（FK），其中包含实际问题。

我想要获得的是一个问题的最高百分比，然后按此顺序获得超过90％（查询）的任何内容。所以......让我们说有100个人被问到“你快乐吗？＆＃39;题。 5％表示否，95％表示是。因为它是95％，它会出现在查询中。另一方面，如果20％的人回答没有，你认为自己聪明吗？＆＃39;问题，80％表示是，然后它就不会显示出来。我想我需要接下来一堆子查询，但我不确定。这就是我所拥有的......但我知道我已经走了。任何帮助将不胜感激。

SELECT question, answer, round(COUNT(*)/(select count(*) from questions 
INNER JOIN questions_only on questions_only.question_ID=questions.question_ID)*100) AS    
'Percent' FROM questions 
INNER JOIN answers_only ON answers_only.answer_ID=questions.answer_ID 
INNER JOIN questions_only ON questions_only.question_ID=questions.question_ID 
GROUP BY answer order by COUNT(*) DESC

这里的最终目标是找出所有容易回答的问题，大多数人都赞成。

                         wanted results from query
╔═══════════════════════════════╦══════════════════╦════════════════╗
║ question                      ║      answer      ║      Percent   ║ 
╠═══════════════════════════════╬══════════════════╬════════════════╬
║  Are you happy?               ║          Yes     ║          97    ║        
║  You think you're smart?      ║          Yes     ║          96    ║          
║  1-5 How exciting is surfing? ║            5     ║          92    ║          
╚═══════════════════════════════╩══════════════════╩════════════════╝

Answer 1

您应该为每个问题创建一个包含详细答案的子查询，并将其与您的QUESTIONS表联系起来：

SELECT MAX(questions_only.question),
       MAX(answers_only.answer),
       (COUNT(*)/MAX(t.all_count))*100 as answer_percent


FROM questions
   JOIN ( 
           SELECT question_id, 
                  COUNT(*) as all_count
           FROM questions
           GROUP BY question_id 
         ) as t on questions.question_ID=t.question_ID
   JOIN questions_only ON questions.question_ID=questions_only.question_ID
   JOIN answers_only ON questions.answer_id=answers_only.answer_id

GROUP BY questions.question_ID,questions.answer_ID
HAVING (COUNT(*)/MAX(t.all_count))*100>=90

SQLFiddle demo

Answer 2

我没有在工作室检查这个，但它应该给你一种方法。

select question, answer,  round((c/c1)*100) from 
   (select question, answer, count (*) c1, c from
      questions q 
      join questions_only qo on q.question_ID = qo.question_ID
      join answers a on q.answer_ID = a.answer_ID
      join (select question_ID, count(*) c fromquestions ) 
      as q1 on q.question_ID = q1.question_ID
      Group by question, answer) 
   as table

从sql查询和分组中获取百分比

2 个答案: