mysql标准差

[shandongperson]

select
pow(sns_examdetail.SCORE-avg(sns_examdetail.SCORE),2) as score_pow ,
sum(pow(sns_examdetail.SCORE-avg(sns_examdetail.SCORE),2)) as sum_pow
from
sns_examdetail,sns_exam,ctb_papers,ctb_subject
where
sns_examdetail.EXAM_ID = sns_exam.EXAM_ID
and
sns_exam.PAPER_ID = ctb_papers.PAPER_ID
and
sns_examdetail.SUBJECT_ID = ctb_subject.SUBJECT_ID
and
ctb_subject.SUBJECT_ID = ctb_papers.SUBJECT_ID
and
sns_examdetail.PROBLEM_ID is null
and
sns_examdetail.STATUS <> 0
and
sns_exam.EXAM_NAME = 'XXXXXXXXX'
and
sns_exam.AREA_ID = '50'
group by subject_name
为什么select中的前半句话可以取得结果，而后半句话就说Invalid use of group function？郁闷的要死

[wwwwb]

必须用聚合函数
max(pow(sns_examdetail.SCORE-avg(sns_examdetail.SCORE),2)) as score_pow

[wwwwb]

必须用聚合函数
max(pow(sns_examdetail.SCORE-avg(sns_examdetail.SCORE),2)) as score_pow

[ACMAIN_CHM]

sum(pow(sns_examdetail.SCORE-avg(sns_examdetail.SCORE),2)) as sum_pow
这句语法不对，对已经AVG了，怎么再对它进行SUM？同样去MAX也是错误的。

[ACMAIN_CHM]

另外如果你是求标准差，你可以直接使用 STD() 函数

select subject_name,STD(SCORE) from ,... group by subject_name

[iihero_]

直接使用STD分析函数