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Write a SQL query to get a list of tenants who are renting more than one apartment.
-- TABLE Apartments
+-------+------------+------------+ | AptID | UnitNumber | BuildingID | +-------+------------+------------+ | 101 | A1 | 1 | | 102 | A2 | 2 | | 103 | A3 | 3 | | 201 | B1 | 4 | | 202 | B2 | 5 | +-------+------------+------------+
-- TABLE Buildings
+------------+-----------+---------------+---------------+ | BuildingID | ComplexID | BuildingName | Address | +------------+-----------+---------------+---------------+ | 1 | 11 | Eastern Hills | San Diego, CA | | 2 | 12 | East End | Seattle, WA | | 3 | 13 | North Park | New York | | 4 | 14 | South Lake | Orlando, FL | | 5 | 15 | West Forest | Atlanta, GA | +------------+-----------+---------------+---------------+
-- TABLE Tenants
+----------+------------+ | TenantID | TenantName | +----------+------------+ | 1000 | Zhang San | | 1001 | Li Si | | 1002 | Wang Wu | | 1003 | Yang Liu | +----------+------------+
-- TABLE Complexes
+-----------+---------------+ | ComplexID | ComplexName | +-----------+---------------+ | 11 | Luxuary World | | 12 | Paradise | | 13 | Woderland | | 14 | Dreamland | | 15 | LostParis | +-----------+---------------+
-- TABLE AptTenants
+----------+-------+ | TenantID | AptID | +----------+-------+ | 1000 | 102 | | 1001 | 102 | | 1002 | 101 | | 1002 | 103 | | 1002 | 201 | | 1003 | 202 | +----------+-------+
-- TABLE Requests
+-----------+---------+-------+-------------+ | RequestID | Status | AptID | Description | +-----------+---------+-------+-------------+ | 50 | Rented | 101 | | | 60 | Pending | 103 | | +-----------+---------+-------+-------------+
这道题让我们租了不止一间公寓的人，那么我们需要两个表Tenants和AptTenants，其他的表都不需要，那么我们可以用Inner Join来关联两个表，关于SQL的各种Join请参见我之前的博客SQL Left Join, Right Join, Inner Join, and Natural Join 各种Join小结，然后我们还需要用Group by和Count关键字来表示在AptTenants表中出现的次数大于1的TenantID，然后在Tenants表中找到名字返回：
解法一：
SELECT TenantName FROM Tenants INNER JOIN (SELECT TenantID FROM AptTenants GROUP BY TenantID HAVING COUNT(*) > 1) C ON Tenants.TenantID = C.TenantID;
下面这种解法用了Using关键字指定了相同列TenantID：
解法二：
SELECT TenantName FROM Tenants INNER JOIN (SELECT TenantID FROM AptTenants GROUP BY TenantID HAVING COUNT(*) > 1) C USING (TenantID);
[CareerCup] 15.1 Renting Apartment 租房
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原文地址：http://www.cnblogs.com/grandyang/p/5349036.html
