Question

我有以下客户表

CustomerId     CustomerName   CustomerEmail
1              AAA            AAA@AAA.com
2              BBB            BBB@BBB.BBB

这是我展示的示例数据。像这样我在customer表中有50000名客户，现在他们的发票也存储在invoice表中

InvoiceId      InvoiceAmount   InvoiceCustomerID
1              22.00           1
2              58.00           2
3              21.00           2
4              45.00           2

现在，我想在显示客户列表时计算每个客户的发票，例如：

CustomerId     CustomerName   CustomerEmail  TotalInvoice
1              AAA            AAA@AAA.com    1
2              BBB            BBB@BBB.BBB    3

我使用了以下查询，但没有工作

SELECT customer.*,COUNT(invoice.InvoiceId) FROM `customer` as customer INNER JOIN invoice as invoice ON customer.CustomerId= invoice.InvoiceCustomerID

可以建议任何方法来解决这个问题吗？

此外还有5个缺少发票，因此需要花费很多时间才能加载。

Answer 1

SELECT a.CustomerID, a.CustomerName, a.CustomerEmail,
SUM(CASE WHEN b.InvoiceID IS NOT NULL THEN 1 ELSE 0 END) as TotalInvoice
FROM `customer` a LEFT JOIN invoice b ON a.CustomerId= b.InvoiceCustomerID
GROUP BY a.CustomerId, a.CustomerName, a.CustomerEmail

只需使用GROUP BY 此外，您是否需要显示没有发票的客户？如果你这样做，你将不得不把它变成左连接。

Answer 2

您不能使用没有GROUP BY的分组功能：

SELECT customer.CustomerID, customer.Customer.Name, customer.CustomerEmail
    ,COUNT(invoice.InvoiceId) AS RC
FROM `customer` AS customer
INNER JOIN invoice AS invoice ON customer.CustomerId = invoice.InvoiceCustomerID
GROUP BY  customer.CustomerID, customer.Customer.Name, customer.CustomerEmail

Answer 3

select c.CustomerId "CustomerId", 
c.CustomerName "CustomerName", 
c.CustomerEmail "CustomerEmail", 
count(i.InvoiceCustomerID) "TotalInvoice"  
from customer as c 
inner join invocie i 
on(i.InvoiceCustomerID = c.CustomerId) group by
c.CustomerId, c.CustomerName, c.CustomerEmail order by CustomerId

Answer 4

由于批量数据，其他查询需要花费很长时间才能加载

所以我使用以下查询解决了这个问题

SELECT c.CustomerId,c.CustomerName,totalinvoice AS ttlinv FROM customer 
as c LEFT JOIN (select COUNT(InvoiceId) as totalinvoice,InvoiceCustomerID 
from invoice GROUP by InvoiceCustomerID) as io on io.InvoiceCustomerID=c.CustomerId

如何在SQL中统计客户的子数据

4 个答案: