MySQL的投票系统选择

我有一个投票系统以及表如下所示：MySQL的投票系统选择

answers 
----------- 
int id 
text content 

results 
----------- 
int id 
int answers_id

我想选择这样我就可以看到答案的数目和响应，即使没有投票（结果）。我想是这样的：

SELECT 
count(results.answers_id), answers.content 
from answers 
left outer join results 
on answers.id = results.answers_id 
group by results.answers_id

但它不会与0票（结果）的答案。

来源

2015-03-13 meska

COUNT（列）不计NULL（无投票权）。使用count（*）来计算所有 - 包括null！ – jarlh 2015-03-13 09:48:51

这应该做的工作：

SELECT count(results.answer_id), answers.content 
from answers left outer join results on answers.id = results.answer_id 
group by answers.content;

来源

2015-03-13 09:29:10 kubica

尝试

SELECT 
count(result.answers_id) as acount, answers.content 
from answers 
left inner join results 
on answers.id = acount

来源

2015-03-13 09:19:36

我不认为如果这将起作用，不应该添加group by子句，如果在select子句中使用聚合函数？ – 2015-03-13 09:20:53

使用子查询

SELECT a.*, 
    (SELECT COUNT(*) 
    FROM results r 
    WHERE a.id = r.answers_id 
    ) 
FROM answers a

来源

2015-03-13 09:22:37 Javaluca

试试这个，对答案的标识分组。

SELECT 
COUNT(results.answers_id) AS NumberOfAnswers, answers.content 
from answers 
left join results 
on answers.id = results.answers_id 
group by answers.id

要确保你不要有两行内容相同，那么group by answers.content

SELECT 
    COUNT(results.answers_id) AS NumberOfAnswers, answers.content 
    from answers 
    left join results 
    on answers.id = results.answers_id 
    group by answers.content

来源

2015-03-13 09:23:04

使用Ifnull，也是你的GROUP BY语句将其更改为answers.content

SELECT 
IFNULL(count(results.answers_id),0), answers.content 
from answers 
left outer join results 
on answers.id = results.answers_id 
group by answers.content

来源

2015-03-13 09:23:57 Backtrack

我想解决方案是使用IFNULL

SELECT IFNULL(COUNT(r.answers_id), 0), 
     a.content 
FROM answers AS a 
LEFT JOIN results AS r ON a.id = r.answers_id 
GROUP BY r.answers_id

来源

2015-03-13 09:24:58

MySQL的投票系统选择

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