如何把SQL查询到SQL函数

Question

所以我有一个计算之间的两个用户的评价Pearson相关下面的SQL代码：

select @u1avg:=avg(user1_rating), 
    @u2avg:=avg(user2_rating), 
    @u1sd:=stddev(user1_rating), 
    @u2sd:=stddev(user2_rating) 
    from 
(select r1.userId as User1_id,r1.rating as User1_rating, 
     r2.userId as User2_id,r2.rating as User2_rating 

from mydb.ratings r1 join mydb.ratings r2 on r1.itemId = r2.itemid 
where r1.userId=1 and r2.userId=2) sample; 


select (1/(count(r1.rating-1)))*sum(((r1.rating-@u1avg)/@u1sd)*((r2.rating-@u2avg)/@u2sd))*(count(r1.rating)/(1+count(r1.rating))) 

from mydb.ratings r1 join mydb.ratings r2 on r1.itemId = r2.itemid 
where r1.userId=1 and r2.userId=2; 

我愿把它转换成一个函数，例如科尔（A ，B） 任何帮助将是有益的。

我得到的问题是，它说：样品说不准或类似的东西，但是如果我删除样品我得到一个错误说每个表都必须有别名。

Answer 1

我觉得你可以不设在第一个查询派生表，会看到过那个特定的错误 -

SELECT 
    @u1avg:=avg(r1.rating), 
    @u2avg:=avg(r2.rating), 
    @u1sd:=stddev(r1.rating), 
    @u2sd:=stddev(r2.rating) 
FROM mydb.ratings r1 
INNER JOIN mydb.ratings r2 
    ON r1.itemId = r2.itemId 
WHERE r1.userId=1 
AND r2.userId=2; 

SELECT (1/(COUNT(r1.rating-1)))*SUM(((r1.rating-@u1avg)/@u1sd)*((r2.rating-@u2avg)/@u2sd))*(COUNT(r1.rating)/(1+COUNT(r1.rating))) 
FROM mydb.ratings r1 
INNER JOIN mydb.ratings r2 
    ON r1.itemId = r2.itemid 
WHERE r1.userId=1 
AND r2.userId=2;