MySQL左连接重复

Question

表A：qa_invoices

    invoice_code   invoice_date
    1              25
    2              25
    3              26
    4              27

表B：qa_invoicesitems

   invoice_code   item_quantity    item_name
   1              2                Mouse
   1              1                CPU
   2              5                Keyboard
   3              1                Router
   4              3                Switch

我的查询：

SELECT SUM(qa_invoicesitems.item_quantity) AS item_count, COUNT(*) AS invoice_count    
FROM qa_invoices
LEFT OUTER JOIN qa_invoicesitems USING (invoice_code) 
WHERE qa_invoices.invoice_date = 25

我的结果：

item_count  invoice_count
8           3           

我有时间和时间寻找解决方案，问题是计算发票（invoice_count）。

正确的结果是：

item_count  invoice_count
8           2

如果有人可以帮助我解决我的问题。

Answer 1

尝试将DISTINCT添加到您的COUNT。

SELECT SUM(qa_invoicesitems.item_quantity) AS item_count, COUNT(DISTINCT(qa_invoicesitems.invoice_code)) AS invoice_count    
FROM qa_invoices
LEFT OUTER JOIN qa_invoicesitems USING (invoice_code) 
WHERE qa_invoices.invoice_date = 25