Question

我不知道怎么用mysql做这个，我只知道如何做基本查询，我想根据匹配显示结果列表，结果将根据答案的相同值显示...答案可能具有1-10的值

+++++++++++++++TABLE++++++++++++

id  | userName | answer1 | answer2 | answer3 | answer4….

10    Jhon       1        1         3          8

11    Anne       1        2         4          8

12    Mike       7        4         5          7

etc…
++++++++++++++++++++++++++++++++++++++++

如果我在查询中发送值，我想检查答案并显示结果排序我的匹配，更多匹配首先...没有匹配最后

所以，如果我发送结果：

answer1=1  answer2=1 answer3=7 answer4=2...

结果应该是（给回id）

10 11 12

Answer 1

您的桌面设计不合适，您应该将其分为用户和问题表。

如果您无法更改表格设计，可以使用此查询解决问题：

select
  id,
  username,
  if(answer1 = :an1, 1, 0) + if(answer2 = :an2, 1, 0) + if(answer3 = :an3, 1, 0) + if(answer4 = :an4, 1, 0) as total
from
  table
order by total desc

<强>更新更好地解决这个问题：

使用实时演示检查SQLFiddle：http://sqlfiddle.com/#!9/6c145/2。

创建用户表

CREATE TABLE users (
    id INT AUTO_INCREMENT PRIMARY KEY,
    username VARCHAR(50) NOT NULL
);

创建问题表

CREATE TABLE questions (
    id INT PRIMARY KEY,
    correct_answer INT NOT NULL
);

创建用户答案表

 CREATE TABLE user_answers (
     user_id INT,
     question_id INT,
     user_answer TINYINT,
     PRIMARY KEY (user_id, question_id),
     FOREIGN KEY (user_id) REFERENCES users (id) ON DELETE NO   ACTION ON UPDATE NO ACTION,
     FOREIGN KEY (question_id) REFERENCES questions (id) ON DELETE NO ACTION ON UPDATE NO ACTION
 );

要检索数据，您可以使用查询：

SELECT
    tmp.id,
    tmp.username,
    sum(tmp.is_correct) as total
FROM (
    SELECT
        users.id,
        users.username,
        IF (questions.correct_answer = user_answers.user_answer, 1, 0) as is_correct
    FROM
        users
        INNER JOIN user_answers on users.id = user_answers.user_id
        INNER JOIN questions on user_answers.question_id = questions.id
) tmp
GROUP BY tmp.id, tmp.username
ORDER BY total desc;

基于列匹配的Mysql查询

1 个答案: