[CareerCup] 15.2 Renting Apartment II 租房之二

 

Write a SQL query to get a list of all buildings and the number of open requests (Requests in which status equals 'Open').

 

-- 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 | | +-----------+--------+-------+-------------+

 

这道题让我们返回所有的building，并标记出来每个building有多少个Open的requests，那么我们首先要计算每个building的Open的request的个数，然后再和Buildings表联合返回对应的BuildingName，因为Requests表里对应的是Apartment和request，而一个Building里可能有很多个Apartment，所以我们先要联合Apartments表和Requests表来计算每个building的Open请求的个数，我们用内交Inner Join来做，通过AptID列来内交Apartments表和Requests表，然后通过BuildingID来群组，并生成一个名为Count的列，然后再用Buildings表和Count列左交，这里需要注意下，如果某个building没有Open请求，那么我们需要返回0，即需要把NULL变为0，在MySQL里面我们用IFNULL函数来做，而SQL Server则用ISNULL，Oracle则用NVL，详细对比可参见这里。参见代码如下：

 

SELECT BuildingName, IFNULL(Count, 0) AS 'Count' FROM Buildings
LEFT JOIN
(SELECT Apartments.BuildingID, COUNT(*) AS 'Count' FROM Requests
INNER JOIN 
Apartments ON Requests.AptID = Apartments.AptID
WHERE Requests.Status = 'Open' GROUP BY Apartments.BuildingID) ReqCounts
ON ReqCounts.BuildingID = Buildings.BuildingID;

 

运行结果：

+---------------+-------+
| BuildingName  | Count |
+---------------+-------+
| Eastern Hills |     1 |
| East End      |     0 |
| North Park    |     0 |
| South Lake    |     1 |
| West Forest   |     1 |
+---------------+-------+

 

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