Question

我有两个表'Customer'和'CustomerCommuncation'，它们有字段：

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



 **CustomerCommunication**
---------------------------------------------------------
|ID     |    CommValue              |    CustomerID     |
| 1     |    abc@gmail.com          |    1              | 
| 2     |    abcd@gmail.com         |    1              | 
| 3     |    9000000000             |    1              | 
| 4     |    abcde@gmail.com        |    1              | 
| 5     |    xyz@gmail.com          |    2              | 
| 6     |    xyzw@gmail.com         |    2              | 
| 8     |    9000000000             |    3              | 
| 9     |    abcdef@gmail.com       |    3              | 
| 10    |    9000000000             |    5              | 
| 11    |    xyz@gmail.com          |    4              |

这两个表可以统称为：可以联系John menon：abc @ gmail.com，abcd @ gmail.com，abcde @ gmail.com 90亿。

两个客户在他们的名字，姓氏和至少一个CommValue是相似的很简单，customerIDs 1,3和5很相似，因为它们具有相同的FirstName，LastName 并分享PhoneNumber 9000000000。

问题：给出组中的customerID列表。我们必须找到CustomerID的联合与集团内的客户类似。

例如，如果我们得到一个组（1,2），那么结果将是（3,4,5）

我编写的存储过程如下：

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

}

还有其他优化方法吗

Answer 1

可能可以简化为单个SQL。

我假设你想要一个类似于每个命名id的id列表（即，1类似于3和5，而2类似于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

关于SQL小提琴的例子： -

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

存储函数中的Mysql查询优化

1 个答案: