Question

MySQL从未成为我的强项。我创建了一个在线应用程序，允许用户为其他人创建测试，只是为了查看他们是否通过。这些测试的数据库结构如下：

user_test
  - id (integer, primary key, auto increment)
  - owner (integer)

user_test_question
 - id (integer, primary key, auto increment)
 - belongs_to_test (integer)
 - question_text (varchar)

user_test_answer
 - id (integer, primary key, auto increment)
 - belongs_to_question (integer)
 - answer_text (integer)
 - answer_correct (integer)

这是我为测试场景创建的架构

CREATE TABLE users (
  user_id INT(11) PRIMARY KEY AUTO_INCREMENT
);

CREATE TABLE user_tests (
  test_id INT(11) PRIMARY KEY AUTO_INCREMENT,
  test_owner INT(11) NOT NULL
);

CREATE TABLE test_questions (
  question_id INT(11) PRIMARY KEY AUTO_INCREMENT,
  question_belongs_to INT(11) NOT NULL,
  question_text VARCHAR(200) NOT NULL
);

CREATE TABLE test_answers (
  answer_id INT(11) PRIMARY KEY AUTO_INCREMENT,
  answer_belongs_to INT(11) NOT NULL,
  answer_text VARCHAR(200) NOT NULL,
  answer_correct SMALLINT(1) DEFAULT 0
);

INSERT INTO users (user_id) VALUES (NULL);

INSERT INTO user_tests(test_id, test_owner) VALUES (NULL, 1);
INSERT INTO test_questions(question_id, question_belongs_to, question_text) VALUES (NULL, 1, "first question");
INSERT INTO test_questions(question_id, question_belongs_to, question_text) VALUES (NULL, 1, "second question");
INSERT INTO test_questions(question_id, question_belongs_to, question_text) VALUES (NULL, 1, "third question");

INSERT INTO test_answers(answer_id, answer_belongs_to, answer_text, answer_correct) VALUES (null, 1, "Question 1 - answer 1", 0);
INSERT INTO test_answers(answer_id, answer_belongs_to, answer_text, answer_correct) VALUES (null, 1, "Question 1 - answer 2", 0);
INSERT INTO test_answers(answer_id, answer_belongs_to, answer_text, answer_correct) VALUES (null, 1, "Question 1 - answer 3", 0);

INSERT INTO test_answers(answer_id, answer_belongs_to, answer_text, answer_correct) VALUES (null, 2, "Question 2 - answer 1", 0);
INSERT INTO test_answers(answer_id, answer_belongs_to, answer_text, answer_correct) VALUES (null, 2, "Question 2 - answer 1", 0);

INSERT INTO test_answers(answer_id, answer_belongs_to, answer_text, answer_correct) VALUES (null, 3, "Question 3 - answer 1", 0);
INSERT INTO test_answers(answer_id, answer_belongs_to, answer_text, answer_correct) VALUES (null, 3, "Question 3 - answer 2", 0);
INSERT INTO test_answers(answer_id, answer_belongs_to, answer_text, answer_correct) VALUES (null, 3, "Question 3 - answer 3", 0);
INSERT INTO test_answers(answer_id, answer_belongs_to, answer_text, answer_correct) VALUES (null, 3, "Question 3 - answer 4", 0);

以下是我尝试使用的查询：

SELECT * FROM `user_tests`, `test_questions`, `test_answers`
WHERE `user_tests`.`test_owner` = 1 #The "user id" 
AND `test_answers`.`answer_id` = `test_questions`.`question_id`
GROUP BY `test_questions`.`question_id` 
ORDER BY `test_questions`.`question_id`

结果只显示每个问题一个答案，预期结果如下：

按问题ID的顺序列出的答案，如下：

question_id: 1, answer_id: 1, text: Question 1 - Answer 1
question_id: 1, answer_id: 2, text: Question 1 - Answer 2
question_id: 1, answer_id: 3, text: Question 1 - Answer 3
question_id: 2, answer_id: 4, text: Question 2 - Answer 1
question_id: 2, answer_id: 5, text: Question 2 - Answer 2
question_id: 3, answer_id: 6, text: Question 3 - Answer 1
question_id: 3, answer_id: 7, text: Question 3 - Answer 2
question_id: 3, answer_id: 8, text: Question 3 - Answer 3
question_id: 3, answer_id: 9, text: Question 3 - Answer 4

Answer 1

您在question_id上使用GROUP BY。这将使每个问题只显示一行，您必须在查询中删除此行

此外，您在查询中还有以下行：

AND `test_answers`.`answer_id` = `test_questions`.`question_id`

这对我来说似乎不正确？这应该是belongs_to_question？|

我在自己的数据库中重建表（sqlfiddle体验超时）：

SELECT * FROM `user_tests`, `test_questions`, `test_answers`
WHERE `user_tests`.`test_owner` = 1 #The "user id" 
AND `test_answers`.`answer_belongs_to` = `test_questions`.`question_id`
ORDER BY `test_questions`.`question_id`, `test_answers`.`answer_id`

answer_id = 5的answer_text虽然不正确，但我想这只是样本数据中的一个拼写错误。

Answer 2

第一个问题是您的GROUP BY不需要。你需要它，例如，如果你想计算好的答案数每个问题

如果您真的想要输出的话，那么您将遇到的第二个问题是ORDER BY。您需要为answer_id

添加第二个参数

如草莓所述，如果使用JOIN语法（显式）而不是多个FROM（隐式），则可读性更好

SELECT question_id, answer_id, ...
FROM user_tests
INNER JOIN ON test_questions.question_belongs_to = user_tests.test_id
...
WHERE user_tests.test_owner = 1

Answer 3

这正是您所寻找的：

var path = require('path');

function isBackgroundFile(file) {
    // check file extension or something.
    if(path.extname(file) === ".png" || path.extname(file) === ".jpg") {
        return true;
    }
    return false;
}

function isNonBackgroundFile(file) {
    return !isBackgroundFile(file);
}

SQL Fiddle

您使用的是fs.readdir，如果您想要所有答案，则不需要。

另外你使用隐式连接语法，这是一个坏习惯，总是使用显式JOIN语法

尝试进行查询以获得测试中所有问题的所有答案

3 个答案: