SUM DISTINCT VALUE [加入

Question

Order_ID 
========= 
id price 
A 10 
A 10 
B 20 
B 20 
C 30 
C 30 
D 40 
D 40 

Client 
================== 
Client Name id 
1 ClientInc. A 
1 ClientInc. A 
1 ClientInc. B 
1 ClientInc. B 
1 ClientInc. C 
1 ClientInc. C 
1 ClientInc. D 
1 ClientInc. D 

我有我需要加入（Order_ID上和客户端），并要总结由不同ORDER_ID价格，并创建下面的报告两个表：

Desired Solution 
========================   
id Name Sum(Price) 
1 ClientInc. 100 

这是我使用当前查询：

SELECT merchant, 
name, 
SUM(price) 
FROM order_id a 
JOIN client b 
ON a.id = b.id 
GROUP BY merchant, name 

它是通过总结每ORDER_ID显示下面的输出，但问题是，我要总结一个独特的命令ID：

Current Wrong Report 
======================  
id Name Sum(Price) 
1 ClientInc. 200 

Answer 1

SELECT merchant, 
     name, 
     SUM(price) 
FROM (SELECT DISTINCT id,price 
     FROM order_id 
    ) a 
JOIN client b 
ON a.id = b.id 
GROUP BY merchant, name;