日期:2014-05-17  浏览次数:20738 次

求这样一个SQL语句
表a结构
  Name BigClass SmallClass
  张三 1 101
  李四 1 102
  王五 2 201
表b结构
  smallClass time1 time2
  101 2012-10-19 2012-10-19
  102 2012-10-19
  201 2012-10-19
  101 2012-10-19 
  101 2012-10-19

要求结果
  姓名 TIME1和TIME2都存在的记录数 只有TIME1的记录数 只有TIME2的记录数 
 

------解决方案--------------------
try



SQL code
--> 测试数据:[ta]
IF OBJECT_ID('[ta]') IS NOT NULL DROP TABLE [ta]
GO 
CREATE TABLE [ta]([Name] VARCHAR(4),[BigClass] INT,[SmallClass] INT)
INSERT [ta]
SELECT '张三',1,101 UNION ALL
SELECT '李四',1,102 UNION ALL
SELECT '王五',2,201

--> 测试数据:[tb]
IF OBJECT_ID('[tb]') IS NOT NULL DROP TABLE [tb]
GO 
CREATE TABLE [tb]([smallClass] INT,[time1] DATETIME,[time2] DATETIME)
INSERT [tb]
SELECT 101,'2012-10-19','2012-10-19' UNION ALL
SELECT 102,'2012-10-19',NULL UNION ALL
SELECT 201,'2012-10-19',NULL UNION ALL
SELECT 101,'2012-10-19',NULL UNION ALL
SELECT 101,'2012-10-19',NULL
--------------开始查询--------------------------

SELECT [Name],
SUM (case WHEN [time1] is NOT NULL AND [time2] is NOT NULL THEN 1 ELSE 0 END) AS [time1 time2 都不为null],
SUM (case WHEN [time1] is NOT NULL AND [time2] is  NULL THEN 1 ELSE 0 END) AS [time2 为null],
SUM (case WHEN [time1] is  NULL AND [time2] is NOT NULL THEN 1 ELSE 0 END) AS [time1 为null]
FROM [ta] AS a JOIN  [tb] AS b  ON a.[smallClass]=b.[smallClass]
GROUP BY [Name]
----------------结果----------------------------
/* 
Name    time1 time2 都不为null    time2 为null    time1 为null
李四    0    1    0
王五    0    1    0
张三    1    2    0
*/