这基本上就是我想要实现的目标;
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
实现这一目标的最佳方法是什么?
谢谢
这基本上就是我想要实现的目标;
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
实现这一目标的最佳方法是什么?
谢谢
任何聚合比较都需要在 Group By 之后进入 HAVING 子句。请记住,根据您放置非聚合条件的位置,您可能会得到不同的结果,因为您将汇总通过 WHERE 包含/排除的内容。
IE
SELECT COUNT(DISTINCT No)
FROM Table
WHERE Type = 1
GROUP BY Week
HAVING COUNT(DISTINCT No) > Num
可能不同于
SELECT COUNT(DISTINCT No)
FROM Table
GROUP BY Week
HAVING COUNT(DISTINCT No) > Num
AND Type = 1
SELECT
COUNT(DISTINCT No)
FROM
Table
WHERE Type = 1 OR Type = 2
GROUP BY
Week
Having COUNT(DISTINCT No) > Num
这是可能的答案。
用这个:
SELECT
COUNT(DISTINCT No) as num1
FROM TableName
WHERE Type IN (1,2)
GROUP BY Week
HAVING num1 > Num
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;
如果您按周分组,您肯定想查看计数是针对哪一周的?!