2014-01-15 219 views
0

我有兩個表「客戶」和「CustomerCommuncation」具有字段:MySQL查詢優化

**Customer** 
--------------------------------------------- 
|ID  | FirstName | LastName  | 
| 1  | John   | Menon   | 
| 2  | george  | cool   | 
| 3  | John   | Menon   | 
| 4  | george  | cool   | 
| 5  | John   | Menon   | 



**CustomerCommunication** 
--------------------------------------------------------- 
|ID  | CommValue    | CustomerID  | 
| 1  | [email protected]   | 1    | 
| 2  | [email protected]   | 1    | 
| 3  | 9000000000    | 1    | 
| 4  | [email protected]  | 1    | 
| 5  | [email protected]   | 2    | 
| 6  | [email protected]   | 2    | 
| 8  | 9000000000    | 3    | 
| 9  | [email protected]  | 3    | 
| 10 | 9000000000    | 5    | 
| 11 | [email protected]   | 4    | 

這兩個表可以被統稱爲: 約翰·梅農已經可以聯繫:ABC @ gmail.com,ABCD @ gmail.com,ABCDE @ gmail.com 9000000000.

兩個客戶是相似時,他們的名字,姓氏和CommValue的ATLEAST一個是 similar.Clearly customerIDs 1,3和5是類似,因爲他們有相同的名字,姓氏 和份額s PhoneNumber 9000000000.

Question: 給定一個客戶ID到一個組的列表。我們必須找到類似於本集團客戶的 的CustomerID聯盟。

例如,如果我們給出一個組(1,2)比的結果將是(3,4,5)

我已經寫存儲過程,如:

CREATE DEFINER= FUNCTION `stroredfunc`(custlist varchar(1000)) RETURNS varchar(10000) 
{ 
begin 
declare v_commvalue varchar(10000) default ''; 
declare v_LastName varchar(1000) default ''; 
declare v_firstname varchar(1000) default ''; 
declare v_result varchar(10000) default ''; 

select group_concat(Distinct c.FirstName SEPARATOR ',') as v_firstname, 
group_concat(Distinct c.LastName SEPARATOR ',') as v_LastName, 
group_concat(Distinct cc.CommValue SEPARATOR ',') as v_commvalue 
from Customer as c inner join CustomerCommunication as cc 
on cc.CustomerID=c.ID where FIND_INSET(c.ID,custlist); 


select group_concat(distinct cc.CustomerID) into v_result from Customer as c inner join 
CustomerCommunication as cc 
on cc.CustomerID=c.ID where FIND_IN_SET(c.FirstName,v_firstname) and 
FIND_IN_SET(c.LastName,v_LastName) and 
FIND_IN_SET(cc.CommunicationValue,v_commvalue) 
; 

return v_result; 
end 

}

有沒有其他優化的方法可以做到這一點

回答

0

可能可以簡化爲一個單一的SQL。

我假定要的類似的ID到每個命名ID的列表(即,1類似於圖3和5以及圖2是類似於4): -

SELECT a.ID, a.FirstName, a.LastName, GROUP_CONCAT(DISTINCT b.Id) 
FROM Customer a 
INNER JOIN Customer b 
ON a.FirstName = b.FirstName 
AND a.LastName = b.LastName 
AND a.Id != b.Id 
INNER JOIN CustomerCommunication c 
ON a.Id = c.CustomerID 
INNER JOIN CustomerCommunication d 
ON a.Id = d.CustomerID 
AND c.CommValue = d.CommValue 
WHERE a.Id IN (1,2) 
GROUP BY a.ID, a.FirstName, a.LastName 

實施例敲起來在SQL小提琴: -

http://www.sqlfiddle.com/#!2/cf054/1

+0

非常感謝你。只是一點點的變化,而不是「a.Id = d.CustomerID」它必須是「b.Id = d.CustomerID」 – satishsingh2230