从两个表中选择并数第三个

Question

我有一个用户，问题和答案表，我想做的是根据他们的用户名从问题和用户表中选择，然后根据问题之间的关系计算答案表中的行数和答案表。请记住，表格的当前状态是：

Questions 表有四列（question_id、topic_id、username、question） Answers 表有两列 (question_id, answer) 用户表有两列（用户名、用户邮件） 我试过的查询

SELECT 
questions.question_id,
questions.username,
questions.question,
userlog.user_mail,
COUNT(answers.answer) as answerCount
FROM
questions
LEFT JOIN answers ON answers.question_id = questions.question_id, 
userlog
WHERE
questions.topic_id = '0d3fb89c012b5af12e1e0'
    AND userlog.username = questions.username

上面的问题是，它只返回一行而不是数据库中的三行。

Answer 1

您的查询存在一些问题。首先，COUNT() 的存在使查询成为聚合查询。如果没有GROUP BY，聚合查询只能生成一行。

另一个问题：您对 USERLOG 有一些 JOIN 混淆。希望每个用户在 USERLOG 中只有一行，否则您最终可能会重复计算答案。

试试这个查询

SELECT questions.question_id,
       questions.username,
       questions.question,
       userlog.user_mail,
       COUNT(answers.answer) as answerCount
  FROM questions
  LEFT JOIN userlog ON questions.username = userlog.username
  LEFT JOIN answers ON answers.question_id = questions.question_id
 WHERE questions.topic_id = '0d3fb89c012b5af12e1e0'
 GROUP BY questions.question_id, questions.username, questions.question, userlog.user_mail
 ORDER BY questions.username, questions.question_id

这应该会产生您需要的多行结果集。

Answer 2

试试这个：

SELECT questions.question_id,  
               questions.username, 
               questions.question, 
               userlog.user_mail, 
               (Select COUNT(answers.answer) where answers.question_id = questions.question_id) as answerCount 
FROM questions 
INNER JOIN userlog ON userlog.username = questions.username
WHERE questions.topic_id = '0d3fb89c012b5af12e1e0'

Answer 3

create table users
(
    userId int auto_increment primary key,
    username varchar(100) not null,
    email varchar (100) not null
);

create table questions
(
    qId int auto_increment primary key,
    topicId varchar(50) not null,
    userId int not null,
    question varchar(1000) not null
);

create table answers
(   
    aId int auto_increment primary key,
    qId int not null,
    answer varchar(1000) not null
);

insert users (username,email) values ('sparky','sp@me.com'),('sarah','sarah@me.com');

truncate table questions;   -- for debugging
insert questions (topicId,userId,question) values ('0d3fb89c012b5af12e1e0',1,'Does life exist outside our galaxy?');
insert questions (topicId,userId,question) values ('0d3fb89c012b5af12e1e0',1,'Are fish really that dumb? Really?');
insert questions (topicId,userId,question) values ('xxxxxx',1,'Am I here?');
insert questions (topicId,userId,question) values ('xxxxxx',1,'Am you here?');

truncate table answers; -- for debugging
insert answers (qId,answer) values (1,'I hope so.');
insert answers (qId,answer) values (1,'I think so.');
insert answers (qId,answer) values (2,'What is wrong with you.');
insert answers (qId,answer) values (2,'Fish are nice.');
insert answers (qId,answer) values (2,'I like turtles.');
insert answers (qId,answer) values (3,'I like turtles too.');
insert answers (qId,answer) values (3,'Me 3.');

-- select * from users;
-- select * from questions;
-- select * from answers;


select 
q.qId,u.username,q.question,u.email,count(a.aId) as AnswerCount
from users u
join questions q
on q.userId=u.userId  and q.topicId='0d3fb89c012b5af12e1e0'
join answers a
on a.qId=q.qId
group by q.qId,u.username,q.question,u.email


+-----+----------+-------------------------------------+-----------+-------------+
| qId | username | question                            | email     | AnswerCount |
+-----+----------+-------------------------------------+-----------+-------------+
|   1 | sparky   | Does life exist outside our galaxy? | sp@me.com |           2 |
|   2 | sparky   | Are fish really that dumb? Really?  | sp@me.com |           3 |
+-----+----------+-------------------------------------+-----------+-------------+
2 rows in set (0.04 sec)