[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         |          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 | Open   |   101 |             |
|        60 | Close  |   103 |             |
|        70 | Close  |   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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