5

我有以下方案：

CREATE TABLE IF NOT EXISTS `answers` (
`id` bigint(20) unsigned NOT NULL,
`answer` varchar(200) NOT NULL,
`username` varchar(15) NOT NULL,
`date` datetime NOT NULL,
PRIMARY KEY (`id`,`username`)
) ENGINE=MyISAM DEFAULT CHARSET=utf8;

INSERT INTO `answers` (`id`, `answer`, `username`, `date`) VALUES
(1, 'gfdsf', 'guy', '2012-12-22 00:00:00'),
(4, 'gfdddsfs', 'maricela', '2012-12-22 00:00:00'),
(4, 'gfddsfs', 'mikha', '2012-12-22 00:00:00'),
(4, 'gfdsfs', 'guy', '2012-12-22 00:00:00');

CREATE TABLE IF NOT EXISTS `questions` (
`id` bigint(20) unsigned NOT NULL AUTO_INCREMENT,
`asker_username` varchar(15) NOT NULL,
`target_username` varchar(15) NOT NULL,
`question` varchar(200) NOT NULL,
`hide` enum('y','n') NOT NULL DEFAULT 'n',
`date` datetime NOT NULL,
PRIMARY KEY (`id`)
) ENGINE=MyISAM  DEFAULT CHARSET=utf8 AUTO_INCREMENT=5 ;

INSERT INTO `questions` (`id`, `asker_username`, `target_username`, `question`, `date`) VALUES
(1, 'mikha', 'guy', 'testo festo', '2012-12-22 00:00:00'),
(2, 'mikha', 'guy', 'saaaaaaaar', '2012-12-22 00:00:00'),
(3, 'sys.tem', 'every.one', 'test g1', '2012-12-06 00:00:00'),
(4, 'sys.tem', 'every.one', 'test g2', '2012-12-06 00:00:00');

我使用以下查询：

   SELECT        
   questions.id AS questionid,
   COUNT(answers.username) AS count_everyone,
   answers.username  
   FROM questions
   LEFT JOIN answers ON questions.id = answers.id
   GROUP BY questions.id,answers.username

问题出在COUNT(answers.username. 我想计算每个问题的答案，但查询显示计数为 1。例如，问题 ID 4 被回答 3 次，但COUNT(answers.username)显示为 1 而不是 3。

这是预期的结果：

         questionid count_everyone  username
               1         1            guy
               2         0            null
               3         0            null
               4         3             guy
               4         3           maricela
               4         3             mikha

这是我实际得到的结果：

         questionid count_everyone  username
               1         1            guy
               2         0            null
               3         0            null
               4         1             guy
               4         1           maricela
               4         1             mikha

谢谢

4

5 回答 5

4 
select q.id, coalesce(j.AnswerCount, 0) as AnswerCount, a.username

from questions q

left outer join

(select id as Qid, count(answer) as AnswerCount
from answers
group by id) j

on q.id = j.Qid

left outer join 
answers a on q.id = a.id
于 2012-12-22T21:28:51.433 回答
2

要获得正确的计数，您应该只按问题 ID 分组，而不是用户名：

SELECT        
    questions.id AS questionid,
    COUNT(answers.username) AS count_everyone
FROM questions
LEFT JOIN answers ON questions.id = answers.id
GROUP BY questions.id

如果您必须在同一查询中获取用户名，请使用连接：

SELECT questionid, count_everyone, username
FROM
(
    SELECT        
        questions.id AS questionid,
        COUNT(answers.username) AS count_everyone
    FROM questions
    LEFT JOIN answers ON questions.id = answers.id
    GROUP BY questions.id
) T1
LEFT JOIN answers ON T1.questionid = answers.id

sqlfiddle

或GROUP_CONCAT：

SELECT        
    questions.id AS questionid,
    COUNT(answers.username) AS count_everyone,
    GROUP_CONCAT(answers.username) AS usernames
FROM questions
LEFT JOIN answers ON questions.id = answers.id
GROUP BY questions.id

sqlfiddle

或相关子查询：

SELECT        
    questions.id AS questionid,
    (SELECT COUNT(*) FROM answers WHERE questions.id = answers.id) AS count_everyone,
    answers.username
FROM questions
LEFT JOIN answers ON questions.id = answers.id

sqlfiddle

于 2012-12-22T21:03:39.233 回答
1

编辑

我想你在找这个，它会显示所有的问题 id 并计算所有的答案，并显示所有的用户名。

演示 SLQFIDDLE

于 2012-12-22T21:16:03.367 回答
1

这个怎么样：

SELECT        
  questions.id AS questionid,
  COUNT(answers.username) AS count_everyone,
  GROUP_CONCAT(answers.username) users
FROM questions
LEFT JOIN answers ON questions.id = answers.id
WHERE questions.target_username = 'every.one'
GROUP BY questions.id

小提琴

于 2012-12-22T22:17:17.700 回答
0

据我所知，你应该改变这个：

COUNT(answers.username) AS count_everyone,

至：

COUNT(answers.id) AS count_everyone,

因为您实际上想计算具有相同 ID 的答案的数量...

于 2012-12-22T21:02:58.747 回答