MySQL INNER JOIN-面试得分的问答

Question

我正在处理这个面试候选人的小项目，并给他们的答案打分。

例如：

问题（A）带有类似的答案：

a1 - Alternative 01 (02 points)
a2 - Alternative 02 (05 points)
a3 - Alternative 03 (00 points)

然后我创建的数据结构如下：

访问此数据模式here

我想做什么？

[使用PHP创建一个接收类似JSON的API：

"query":{
     "candidate":"JOHN",
     "answer":"a1, a2"
     "question_id": 1
}

然后将INSERT数据发送到候选表

然后需要TABLE至JOIN

已回答的问题并求职者选择的总和得分

我正在尝试做，但是没有得到结果。

感谢您阅读到现在为止。

Answer 1

创建表的SQL代码：

CREATE TABLE `Candidates` (
  `id` int PRIMARY KEY AUTO_INCREMENT,
  `name` text,
  `age` int,
  `birthday` date
);

CREATE TABLE `Candidate_Answers` (
  `id` int PRIMARY KEY AUTO_INCREMENT,
  `Candidate_id` int,
  `Question_id` int,
  `Interview_id` int,
  `Candidate_Answer` text
);

CREATE TABLE `Candidate_Score` (
  `Candidate_Answer_id` int PRIMARY KEY,
  `Answers_Score_id` int,
  `Candidate_Score` int
);

CREATE TABLE `Questions` (
  `id` int PRIMARY KEY AUTO_INCREMENT,
  `Question_Type_id` int
);

CREATE TABLE `Type_of_Question` (
  `id` int PRIMARY KEY AUTO_INCREMENT,
  `Type` varchar(1)
);

CREATE TABLE `Interview` (
  `id` int PRIMARY KEY AUTO_INCREMENT,
  `date` date
);

CREATE TABLE `Questions_in_Interview` (
  `Interview_id` int,
  `Question_id` int,
  PRIMARY KEY (`Interview_id`, `Question_id`)
);

CREATE TABLE `Answers_Score` (
  `id` int PRIMARY KEY,
  `Question_id` int,
  `Score` int
);

ALTER TABLE `Candidate_Answers` ADD FOREIGN KEY (`Candidate_id`) REFERENCES `Candidates` (`id`);

ALTER TABLE `Candidate_Answers` ADD FOREIGN KEY (`Interview_id`) REFERENCES `Questions_in_Interview` (`Interview_id`);

ALTER TABLE `Candidate_Score` ADD FOREIGN KEY (`Candidate_Answer_id`) REFERENCES `Candidate_Answers` (`id`);

ALTER TABLE `Candidate_Score` ADD FOREIGN KEY (`Answers_Score_id`) REFERENCES `Answers_Score` (`id`);

ALTER TABLE `Answers_Score` ADD FOREIGN KEY (`Question_id`) REFERENCES `Questions` (`id`);

ALTER TABLE `Questions_in_Interview` ADD FOREIGN KEY (`Question_id`) REFERENCES `Questions` (`id`);

ALTER TABLE `Questions_in_Interview` ADD FOREIGN KEY (`Interview_id`) REFERENCES `Interview` (`id`);

ALTER TABLE `Candidate_Answers` ADD FOREIGN KEY (`Question_id`) REFERENCES `Questions_in_Interview` (`Question_id`);

ALTER TABLE `Questions` ADD FOREIGN KEY (`Question_Type_id`) REFERENCES `Type_of_Question` (`id`);