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 | +---------------+-------+
本文转自博客园Grandyang的博客，原文链接：租房之二[CareerCup] 15.2 Renting Apartment II ，如需转载请自行联系原博主。
