[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 | +---------------+-------+
