[CareerCup] 15.1 Renting Apartment 租房

 

Write a SQL query to get a list of tenants who are renting more than one apartment.

 

-- TABLE Apartments

+-------+------------+------------+ | AptID | UnitNumber | BuildingID | +-------+------------+------------+ | 101 | A1 | 11 | | 102 | A2 | 12 | | 103 | A3 | 13 | | 201 | B1 | 14 | | 202 | B2 | 15 | +-------+------------+------------+

 

-- TABLE Buildings

+------------+-----------+---------------+---------------+ | BuildingID | ComplexID | BuildingName | Address | +------------+-----------+---------------+---------------+ | 11 | 1 | Eastern Hills | San Diego, CA | | 12 | 2 | East End | Seattle, WA | | 13 | 3 | North Park | New York | | 14 | 4 | South Lake | Orlando, FL | | 15 | 5 | West Forest | Atlanta, GA | +------------+-----------+---------------+---------------+

 

-- TABLE Tenants

+----------+------------+ | TenantID | TenantName | +----------+------------+ | 1000 | Zhang San | | 1001 | Li Si | | 1002 | Wang Wu | | 1003 | Yang Liu | +----------+------------+

 

-- TABLE Complexes

+-----------+---------------+ | ComplexID | ComplexName | +-----------+---------------+ | 1 | Luxuary World | | 2 | Paradise | | 3 | Woderland | | 4 | Dreamland | | 5 | LostParis | +-----------+---------------+

 

-- TABLE AptTenants

+----------+-------+ | TenantID | AptID | +----------+-------+ | 1000 | 102 | | 1001 | 102 | | 1002 | 101 | | 1002 | 103 | | 1002 | 201 | | 1003 | 202 | +----------+-------+

 

-- TABLE Requests

+-----------+--------+-------+-------------+ | RequestID | Status | AptID | Description | +-----------+--------+-------+-------------+ | 50 | Open | 101 | | | 60 | Closed | 103 | | | 70 | Closed | 102 | | | 80 | Open | 201 | | | 90 | Open | 202 | | +-----------+--------+-------+-------------+

 

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);

 

运行结果：

+------------+
| TenantName |
+------------+
| Wang Wu    |
+------------+

 

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