Question

我有这个问题：

    SELECT 

    SUM(IF(b.correct = 1,1,0)) correct_answers,
    SUM(IF(b.correct != 1,1,0)) incorrect_answers,
    SUM(IF(a.answer IS NULL) 1,0)) as no_answers,
    a.ID
    FROM `a` join b on a.answer  = b.id 
    where a.created_at between '2014-06-10' and '2014-06-17' 
    group by a.ID order by correct_answers desc

基本上我试图获取每个用户测验的正确，不正确和无答案的数量。所以我得到了正确和错误的计数。但是当我尝试计算无答案的数量时（表a中的答案列为NULL）我得到全部为零。这个查询有什么问题吗？

Answer 1

试试这个：

SELECT SUM(b.correct = 1) correct_answers,
       SUM(b.correct != 1) incorrect_answers,
       SUM(a.answer IS NULL) AS no_answers,
       a.ID
FROM a 
LEFT JOIN b ON a.answer = b.id 
WHERE a.created_at BETWEEN '2014-06-10' AND '2014-06-17' 
GROUP BY a.ID 
ORDER BY SUM(b.correct = 1) DESC;

Answer 2

你错过了一个＆＃39;，＆＃39;在第3行

更改：

 SUM(IF(a.answer IS NULL) 1,0)) as no_answers,

要：

 SUM(IF(a.answer IS NULL, 1,0)) as no_answers,

Answer 3

纠正这个

SUM(IF(a.answer IS NULL), 1,0)) as no_answers

而不是

SUM(IF(a.answer IS NULL) 1,0)) as no_answers

Answer 4

您的查询对于您所描述的内容看起来是正确的。这是一个可能的简化以及left join：

SELECT SUM(b.correct = 1) correct_answers,
       SUM(b.correct <> 1) incorrect_answers,
       SUM(a.answer IS NULL) as no_answers,
       a.ID
FROM `a` left join
      b
      on a.answer  = b.id 
where a.created_at between '2014-06-10' and '2014-06-17' 
group by a.ID
order by correct_answers desc;

MySQL中不需要if语句。

Answer 5

SELECT 
SUM(IF(b.correct = 1,1,0)) correct_answers,
SUM(IF(b.correct != 1,1,0)) incorrect_answers,
SUM(IF(a.answer IS NULL,1,0)) as no_answers,     <-- You missed a comma here. So, it was always evaluating to 0.
a.ID
FROM `a` join b on a.answer  = b.id 
where a.created_at between '2014-06-10' and '2014-06-17' 
group by a.ID order by correct_answers desc

在IF条件下的NULLS计数

5 个答案: