Question

请注意我在代码

下面做了一个SELECT

我创建了一个SQL小提琴，因此您不必在本地数据库中重新创建此示例。 http://sqlfiddle.com/#!9/4c657

问题基本上不是将费率作为数字来获取，而应将其作为百分比，例如：

糟糕：25％，好10％，非常好：30％：优秀：45％

目前我只是在显示数字，我不确定是否应该使用SELECT FROM（SELECT ...）和GROUP BY，或者是否有更好的方法。由于我对此有疑问，我想提出一些建议。谢谢！

由于

    /*What I've got so far*/

    SELECT qg.name AS questionGroup,q.name AS question,(rate=1) AS bad ,(rate=2) AS regular,(rate=3) AS good, (rate=4) AS veryGood, (rate=5) AS excellent from answers AS a INNER JOIN questions AS q ON q.id=a.questionId INNER JOIN questionGroups as qg ON qg.id  = q.groupId;

    CREATE TABLE `guidedVisits` (
      `id` int(10) unsigned NOT NULL AUTO_INCREMENT,
      `date` date NOT NULL,
      PRIMARY KEY (`id`)
    ) ENGINE=InnoDB AUTO_INCREMENT=3 DEFAULT CHARSET=utf8;

    CREATE TABLE `questionGroups` (
      `id` int(10) unsigned NOT NULL AUTO_INCREMENT,
      `name` varchar(20) DEFAULT NULL,
      PRIMARY KEY (`id`)
    ) ENGINE=InnoDB AUTO_INCREMENT=3 DEFAULT CHARSET=utf8;

     CREATE TABLE `questions` (
      `id` int(10) unsigned NOT NULL AUTO_INCREMENT,
      `groupId` int(10) unsigned NOT NULL,
      `name` varchar(120) NOT NULL,
      PRIMARY KEY (`id`),
      KEY `idx_groupId` (`groupId`)
    ) ENGINE=InnoDB AUTO_INCREMENT=5 DEFAULT CHARSET=utf8;

      CREATE TABLE `answers` (
        `id` int(10) unsigned NOT NULL AUTO_INCREMENT,
        `visitId` int(10) unsigned NOT NULL,
        `questionId` int(10) unsigned NOT NULL,
        `rate` tinyint(4) NOT NULL,
        PRIMARY KEY (`id`),
        KEY `idx_questionId` (`questionId`),
        KEY `idx_visitId` (`visitId`)
      ) ENGINE=InnoDB AUTO_INCREMENT=9 DEFAULT CHARSET=utf8;


    INSERT INTO `questionGroups` VALUES (1,'About us'),(2,'Facility');
    INSERT INTO `questions` VALUES (1,1,'Did you like us?'),(2,1,'What do you think about our ads?'),(3,2,'Were our reception chairs comfortable?'),(4,2,'Was everything clean as you would expect?');
    INSERT INTO `guidedVisits` VALUES (1,'2015-03-31'),(2,'2015-03-31');
    INSERT INTO `answers` VALUES (1,1,1,3),(2,1,2,4),(3,1,3,4),(4,1,4,5),(5,1,1,5),(6,1,2,5),(7,1,3,5),(8,1,4,5);

Answer 1

SELECT qg.name AS questionGroup,q.name AS question,  CASE WHEN AVG(rate) < 25 THEN 'bad' WHEN AVG(rate) < 35 THEN 'good' .. END AS rateDesc 
 FROM answers AS a INNER JOIN questions AS q 
   ON q.id=a.questionId INNER JOIN questionGroups as qg 
   ON qg.id  = q.groupId

GROUP BY qg.name，q.name

Answer 2

SELECT r.questionGroup,r.question,r.qid,
FLOOR(AVG(r.bad)*100) AS bad, FLOOR(AVG(r.regular)*100) AS regular,
FLOOR(AVG(r.good)*100) AS good, FLOOR(AVG(r.veryGood)*100) AS veryGood, 
FLOOR(AVG(r.excellent)*100) AS excellent 
FROM (
      SELECT q.id AS qid,qg.id AS gid, 
      qg.name AS questionGroup,q.name AS question,
      (rate=1) AS bad ,(rate=2) AS regular,
      (rate=3) AS good, (rate=4) AS veryGood, 
      (rate=5) AS excellent FROM answers AS a 
      INNER JOIN questions AS q ON q.id=a.questionId 
      INNER JOIN questionGroups as qg ON qg.id  = q.groupId 
      WHERE visitId=1) AS r GROUP BY r.qid;

使用平均值将分数作为一组合格率进行的特定查询

2 个答案: