按发票金额和商品总金额订购客户

Question

我试图按发票金额和发票金额之和订购客户。为了获得发票的价值，我需要对可以（或不可以）在2个表格中的项目进行汇总：产品和支持。结构是这样的：

我尝试了此查询，但不起作用：

SELECT 
    c.id, 
    c.name, 
    COUNT(i.id) as total_amount, 
    SUM(ip.value * ip.quantity) as total_products,
    SUM(sp.value) as total_support
FROM InVoice i
INNER JOIN Costumer c ON i.id_costumer = c.id
LEFT JOIN InVoice_Product ip ON ip.id_invoice = i.id
LEFT JOIN InVoice_Support sp ON sp.id_invoice = i.id
GROUP BY c.id, c.name
ORDER BY total_amount, total_products DESC

如果我在没有LEFT JOINS的情况下执行查询，仅查找发票金额，效果很好。。但是当我尝试添加JOINS时，结果与实际情况不符。

我对此感到困惑。

编辑：

以预期的结果（总计）为准：http://sqlfiddle.com/#!9/f001a7/4

提示完整查询：http://sqlfiddle.com/#!9/f001a7/5

Answer 1

我认为您必须使用DISTINCT计数，并对产品和支持的总和求和。

SELECT 
    c.id, 
    c.name, 
    COUNT(DISTINCT i.id) as invoices_count, 
    SUM(ip.value * ip.quantity) + SUM(is.value) as total_amount
FROM Invoice i
INNER JOIN Costumer c ON i.id_costumer = c.id
LEFT JOIN InVoice_Product ip ON ip.id_invoice = i.id
LEFT JOIN InVoice_Support sp ON sp.id_invoice = i.id
GROUP BY c.id, c.name
ORDER BY invoices_count, total_amount DESC

尝试一下

Answer 2

我认为问题在于对可能为空的值进行数学运算。试试看：

SELECT
    c.id, 
    c.name, 
    COUNT(i.id) as total_amount, 
    SUM(IF(NOT ip.value IS NULL,ip.value * ip.quantity,0)) as total_products,
    SUM(IF(NOT is.value IS NULL,is.value,0)) as total_support
FROM Invoice i
INNER JOIN Costumer c ON i.id_costumer = c.id
LEFT JOIN InVoice_Product ip ON ip.id_invoice = i.id
LEFT JOIN InVoice_Support sp ON sp.id_invoice = i.id
GROUP BY c.id, c.name
ORDER BY total_amount, total_products DESC

Answer 3

只需“联合”两个表以及invoice_product和invoice_support，因为当前查询会在所有三个表（invoice，invoice_product，invoice_support）中找到相同的invoice_id：

select c.id, c.name, count (inv_prod_support.id_invoice), sum(inv_prod_support*quantity) as total_products,
sum(inv_prod_support.value) as total_support from
(select id_invoice, value, 0 as quantity from invoice_support
union select id_invoice, value, quantity from invoice_product ) inv_prod_support, invoice i
inner join customer c on i.id_customer = c.id
inner join inv_prod_support on inv_prod_support.id_invoice = i.id_invoice
group by c.id, c.name
order by total_amount, total_products desc