此代码排序列表和输出为1,但我需要查看列表
SELECT AVG(CustomerID)
FROM Customers
WHERE CustomerID = ( SELECT MIN(CustomerID) FROM Customers )
答案 0 :(得分:1)
根据您的问题,您需要获得AVG
和MIN
值然后比较。
SELECT CustomerID
FROM Customers T
INNER JOIN
(
SELECT MIN(CustomerID) m, AVG(CustomerID) a
FROM Customers
) T2 ON T.CustomerID >=m AND T.CustomerID <= a
答案 1 :(得分:0)
我不明白你为什么要平均customerid 但仍然
select * from customers where customerid <
(SELECT avg(customerid) FROM Customers))