mysql - MySQL AVG()函数返回错误值

Question

我正在尝试计算所选列的平均值。
如果我运行此代码：

SELECT answer_value
FROM survey$question$answer JOIN answer
ON survey$question$answer.answer_letter = answer.answer_letter
WHERE survey$question$answer.survey_id = 12;

我得到这个（这些值是正确的）：


但是，当我运行此命令时：

SELECT AVG(answer_value)
FROM survey$question$answer JOIN answer
ON survey$question$answer.answer_letter = answer.answer_letter
WHERE survey$question$answer.survey_id = 12;

我得到这个：


我得到2的平均值，而5、5和3的正确平均值是4.33
为什么会这样呢？

另一个例子。这段代码在这里：

SELECT answer_value
FROM survey$question$answer JOIN answer
ON survey$question$answer.answer_letter = answer.answer_letter
WHERE survey$question$answer.survey_id = 6;

显示给我：


但是，当我尝试使用此代码计算平均值时：

SELECT AVG(answer_value)
FROM survey$question$answer JOIN answer
ON survey$question$answer.answer_letter = answer.answer_letter
WHERE survey$question$answer.survey_id = 6;

我得到这个：


3、2和1的平均值等于2，而不是9.33

救命？
谢谢！

Answer 1

我认为您错了，您确实在以下其他地方犯了错误，请检查表列的数据类型

select avg( c) from (select 5 as c
union all
select 5
union all
select 3
) a

 c
4.3333

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