在WHERE子句中有COUNT还是HAVING？

Question

这基本上是我试图实现的;

SELECT 
COUNT(DISTINCT No) 
FROM 
Table 
WHERE 
(Type = 1 AND COUNT(DISTINCT No) > Num) 
OR (Type = 2 AND COUNT(DISTINCT No) > Num) 
GROUP BY 
Week 

什么是最好的方法来实现这一目标？

感谢

Answer 1

任何骨料比较需要进入集团后，通过HAVING子句。请记住，根据放置非聚合条件的位置，您可能会得到不同的结果，因为您将通过WHERE包含/排除的内容进行聚合。

即

SELECT COUNT(DISTINCT No) 
FROM Table 
WHERE Type = 1 
GROUP BY Week 
HAVING COUNT(DISTINCT No) > Num 

可以从

SELECT COUNT(DISTINCT No) 
FROM Table 
GROUP BY Week 
HAVING COUNT(DISTINCT No) > Num 
AND Type = 1 

Answer 2

使用此：

SELECT 
COUNT(DISTINCT No) as num1 
FROM TableName 
WHERE Type IN (1,2) 
GROUP BY Week 
HAVING num1 > Num 

Answer 3

SELECT Week, Type, COUNT(DISTINCT No) WeekTally 
    FROM [Table] 
    WHERE Type in (1,2) 
GROUP BY Week, Type 
    HAVING COUNT(DISTINCT No) > @Num 
ORDER BY Week, Type; 

如果你按周分组是不同的，你一定要看到这一周的计数是？