Question

我现在才开始学习SQL，并试图找出这种情况：

我们有3张桌子：

SELECT  CONCAT(IFNULL(Name,''),' ', IFNULL(Surname,'')) as FullName,
        COUNT(purchaseId) as "Quantity of purchases"
FROM Purchases as P
INNER JOIN Clients as C 
on P.ClientID = C.ClientID
GROUP BY C.ClientID,Name, Surname
HAVING COUNT(PurchaseId) > 1;

显示客户购买量（每个客户的总金额）的最佳SQL查询是什么？

它必须仅显示购买超过1次的客户。

结果应包含以下字段：全名，购买数量，总金额。

我有这个查询，但是它只连接两个表。如何同时加入第三张表（Products.Price）并计算每个客户的总金额？

fun <T> Observable<T>.subscribeIf(predicate: () -> Boolean) {
    if (predicate()) {
        subscribe()
    }
}

Answer 1

您可以在下面尝试-

SELECT  CONCAT(IFNULL(Name,''),' ', IFNULL(Surname,'')) as FullName,
        COUNT(purchaseId) as "Quantity of purchases",sum(price) as totalamount
FROM Purchases as P
INNER JOIN Clients as C 
on P.ClientID = C.ClientID
inner join Products p1 on p.productid=p1.productid
GROUP BY C.ClientID,Name, Surname
HAVING COUNT(PurchaseId) > 1

Answer 2

您再次使用INNER JOIN，例如：

SELECT  CONCAT(IFNULL(Name,''),' ', IFNULL(Surname,'')) as FullName,
        COUNT(purchaseId) as "Quantity of purchases"
FROM Purchases as P
INNER JOIN Clients as C 
on P.ClientID = C.ClientID
INNER JOIN Products as PR 
on PR.ProductID = P.ProductID
GROUP BY C.ClientID,Name, Surname
HAVING COUNT(PurchaseId) > 1;

Answer 3

我建议您使用CONCAT_WS()来组合名字。这比您的解决方案更优雅地处理NULL值。

SELECT CONCAT_WS(c.Name, c.Surname) as FullName,
       COUNT(*) as num_purchases,
       SUM(pr.price) as total_price
FROM Clients c INNER JOIN
     Purchases p 
     ON P.ClientID = C.ClientID INNER JOIN
     Products pr
     ON pr.ProductID = p.ProductID
GROUP BY CONCAT_WS(c.Name, c.Surname)
HAVING COUNT(DISTINCT p.PurchaseId) > 1;

请注意COUNT(DISTINCT)子句中的HAVING。这样可以确保客户至少有两个购买。如果您只希望客户购买至少两个产品或购买的商品，则可以使用COUNT(*) －但您的问题是关于购买的。

从3个表中选择具有联接的数据

3 个答案: