Question

我创建了三个表

在这里，我想获取与特定问题（id）相关的所有question_tags和答案。为此需要单个查询或存储过程

create table questions(id varchar(100),title varchar(140),body varchar(2000),
primary key(id));

create table question_tags(id varchar(50),tag_name varchar(50),question_id varchar(100),
primary key(id),foreign key(question_id) references questions(id));


create table answers(id varchar(50),answer varchar(2000),question_id varchar(100),
primary key(id),foreign key(question_id) references questions(id));

以及表格中的以下数据

     mysql> select * from questions;
     +----+-------+-------------+
     | id | title | body        |
     +----+-------+-------------+
     | 1  | a     | hello       |
     | 2  | b     | hii         |
     | 3  | c     | bye         |
     | 4  | d     | how are you |
     +----+-------+-------------+
     4 rows in set (0.03 sec)

     mysql> select * from question_tags
     +----+----------+-------------+
     | id | tag_name | question_id |
     +----+----------+-------------+
     | t1 | java     | 1           |
     | t2 | mysql    | 1           |
     | t3 | jquery   | 1           |
     +----+----------+-------------+
     3 rows in set (0.00 sec)

     mysql> select * from answers;
     +----+-----------+-------------+
     | id | answer    | question_id |
     +----+-----------+-------------+
     | a1 | good      | 1           |
     | a2 | excellent | 1           |
     | a3 | ok        | 1           |
     +----+-----------+-------------+
     3 rows in set (0.00 sec)

Answer 1

我认为您搜索JOIN [ON]语句。对于你的Tablelayout，它应该是这样的：

SELECT `question_tags`.*, `answers`.* 
FROM `questions`
   JOIN  `answers` 
        ON `questions`.`id` = `answers`.`question_id`
   JOIN `question_tags` 
        ON `questions`.`id` = `question_tags`.`question_id`
WHERE `questions`.`id` = {YOUR_ID}

或仅问题ID，答案，标签：

SELECT `questions`.`id` AS `question_id`, 
         `question_tags`.`tag_name`, 
         `answers`.`answer` 
FROM `questions`
   JOIN  `answers` 
        ON `questions`.`id` = `answers`.`question_id`
   JOIN `question_tags` 
        ON `questions`.`id` = `question_tags`.`question_id`
WHERE `questions`.`id` = 0

修改对于一行中的所有事情：

SELECT `questions`.`id` AS `question_id` , ( SELECT GROUP_CONCAT( `question_tags`.`tag_name` SEPARATOR ';') FROM `question_tags` WHERE `question_id` =0 ) AS `Tags` , ( SELECT GROUP_CONCAT( `answers`.`answer` SEPARATOR ';' ) FROM `answers` WHERE `question_id` =0 ) AS `Answers` FROM `questions` WHERE `questions`.`id` =0

请注意，不是这样做的最佳方法。因为我不熟悉MySQL中的SP，所以我无法为您提供存储过程。在我的意见中，没有办法选择/返回/显示您想要的格式作为表格。

Answer 2

查询是：

select * from questions join question_tags on (questions.id = question_tags.question_id) join answers on (questions.id=answers.question_id);

Here是一个小提琴;如果你提供一个问题，它可以帮助我们请阅读有关mySQL命名约定（有关表格的表格）和关键引用（我已将您的更改为integer，但您也可以auto_increment等）。

用于获取数据的单个查询或存储过程

2 个答案: