三个表，两个连接，结果只需要一个表

Question

我有三个表：Clients、Services 和 Client Locations。我正在运行一个查询，该查询需要返回接收特定服务的客户端的位置。所以在 SELECT 中使用第二个表，在 WHERE 中使用第三个表。我正在使用两个 LEFT JOIN 并以不合需要的方式重复我的结果。

这是三个表的简化版本...

客户（客户）

id_client | clientName
----------------------
1         | Abby
2         | Betty
3         | Cathy

Client Services（服务）仅在 WHERE 语句中使用

id_client | date      | serviceType
-----------------------------------
1         | 1/5/2015  | Counseling
1         | 1/12/2015 | Counseling
1         | 1/19/2015 | Counseling
2         | 1/21/2015 | Sup. Group

客户位置（locations）只在SELECT语句中使用

id_client | city
----------------------
1         | Boston, MA
3         | Providence, RI

这里是查询

SELECT clients.clientName,locations.city
FROM clients
LEFT JOIN locations ON clients.id_client=locations.id_client
LEFT JOIN services ON clients.id_client=services.id_client
WHERE services.serviceType='Counseling'

结果

clientName | city
-----------------------
Abby       | Boston, MA
Abby       | Boston, MA
Abby       | Boston, MA

所以这让我 Abby 住在波士顿 3 次，而不是想要的那次。

现在，我确切地知道为什么会发生这种情况。用于服务表的 LEFT JOIN 正在用于结果，并且 Abby 的 3 次咨询会导致该城市重复 3 次。

是否有另一种方法来执行此 JOIN 以便服务表不会导致这样的重复？我已经尝试过 INNER JOIN 并得到了同样的结果。

Answer 1

使用exists:

SELECT c.clientName, l.city
FROM clients c JOIN
     locations l
     ON c.id_client = l.id_client
WHERE EXISTS (SELECT 1
              FROM services s
              WHERE c.id_client = s.id_client AND
                    s.serviceType = 'Counseling'
             );

虽然您可以使用group by 或distinct，但这种方法应该会执行得更好。不需要生成重复的结果，只是为了在另一个步骤中删除它们。

Answer 2

要么使用distinct

SELECT DISTINCT clients.clientName,locations.city
FROM clients
LEFT JOIN locations ON clients.id_client=locations.id_client
LEFT JOIN services ON clients.id_client=services.id_client
WHERE services.serviceType='Counseling'

或group by

SELECT clients.clientName,locations.city
FROM clients
LEFT JOIN locations ON clients.id_client=locations.id_client
LEFT JOIN services ON clients.id_client=services.id_client
WHERE services.serviceType='Counseling'
GROUP BY clients.clientName,locations.city

或子查询

SELECT clients.clientName,locations.city
FROM clients
LEFT JOIN locations ON clients.id_client=locations.id_client
LEFT JOIN (
  SELECT id_client, serviceType 
  FROM services 
  GROUP BY id_client, serviceType 
) services ON clients.id_client=services.id_client
WHERE services.serviceType='Counseling'
GROUP BY clients.clientName,locations.city

Sample SQL Fiddle

Answer 3

您可以获取该服务类型的不同客户端 ID，然后加入客户端和位置以获取有关客户端的更多详细信息。

SELECT clients.clientName,locations.city
FROM 
(Select distinct id_client from services WHERE services.serviceType='Counseling') s
INNER JOIN clients ON clients.id_client = s.id_client
LEFT JOIN locations ON clients.id_client=locations.id_client

Answer 4

您正在汇总详细结果集。因此，请使用 DISTINCT。

SELECT DISTINCT clients.clientName, locations.city
  FROM clients
  LEFT JOIN locations ON clients.id_client=locations.id_client
  LEFT JOIN services ON clients.id_client=services.id_client
 WHERE services.serviceType='Counseling'

或者使用 GROUP BY 查询并提供一些汇总统计信息：

SELECT DISTINCT clients.clientName, locations.city,
       COUNT(*) service_count
  FROM clients
  LEFT JOIN locations ON clients.id_client=locations.id_client
  LEFT JOIN services ON clients.id_client=services.id_client
 WHERE services.serviceType='Counseling' 
 GROUP BY clients.clientName, locations.city

Answer 5

您可以简单地使用 distinct 子句来避免双重结果。

SELECT distinct clients.clientName,locations.city
FROM clients
LEFT JOIN locations ON clients.id_client=locations.id_client
LEFT JOIN services ON clients.id_client=services.id_client
WHERE services.serviceType='Counseling'