如何統計每行的連接中的所有實例？

Question

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'); 

我用下面的查詢：

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 

的問題是與COUNT(answers.username。我想爲每個問題計算答案，但查詢將計數顯示爲1.例如，問題ID 4回答了3次，但COUNT(answers.username)顯示爲1而不是3.

這是預期結果：

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

這是結果我居然得到：

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

感謝

Answer 1

爲了得到正確的計數，你應該僅由問題ID組，但沒有用戶名：

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 

如果有來獲取用戶名S IN相同的查詢，然後使用一個連接：

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

或者相關子查詢：

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

Answer 2

至於我看你應該改變這樣的：

COUNT(answers.username) AS count_everyone, 

到：

COUNT(answers.id) AS count_everyone, 

，你居然要算具有相同ID答案的數目......

Answer 3

編輯

我想你找這個，它會顯示所有問題的ID和計數所有的答案，並顯示所有用戶名。

DEMO SLQFIDDLE

Answer 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 

Answer 5

如何：

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 

FIDDLE