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  • 连续数字区间问题

    这个问题来自于某论坛的SQL Server板块,感觉也是比较典型的问题。在这里我把问题进行简化,并给出2种方案,作为备忘。

    问题描述

    有表Table_Num,如下:

    Num
    4
    5
    6
    10
    11
    12
    13
    20

    通过一个查询返回连续数字区间,结果集如下:

    StartNum EndNum IntervalNums
    4 6 3
    10 13 4
    20 20 1

    解决方案1

    select
    	Num1.Num as StartNum,
    	min(Num2.Num) as EndNum,
    	min(Num2.Num) - Num1.Num + 1 as IntervalNums
    from
    	Table_Num Num1
    	inner join
    	Table_Num Num2
    	on
    		Num1.Num <= Num2.Num
    where
    	(Num1.Num - 1) not in (select Num from Table_Num)
    	and
    	(Num2.Num + 1) not in (select Num from Table_Num)
    group by
    	Num1.Num

    其中where条件找到了Table_Num中的边界数值,group by和min(Num2.Num)保证了数值的连续性。

    解决方案2

    select
    	Num1.Num as StartNum,
    	Num2.Num as EndNum,
    	Num2.Num - Num1.Num + 1 as IntervalNums
    from
    	Table_Num Num1
    	inner join
    	Table_Num Num2
    	on
    		Num1.Num <= Num2.Num
    	inner join
    	Table_Num Num3
    	on
    		Num3.Num between Num1.Num and Num2.Num
    where
    	(Num1.Num - 1) not in (select Num from Table_Num)
    	and
    	(Num2.Num + 1) not in (select Num from Table_Num)
    group by
    	Num1.Num,
    	Num2.Num
    having
    	count(Num3.Num) = Num2.Num - Num1.Num + 1

    个人认为解决方案2比较容易理解,Num1和Num2定位边界数值,Num3为介于Num1和Num2之间的所有数值,而数值的连续性是由having子句count(Num3.Num) = Num2.Num – Num1.Num + 1来保证的。

    除了以上两种方法,应该还有其他方法,大家可以一起讨论。上述两段查询在SQL Server 2005中测试通过。

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  • 原文地址:https://www.cnblogs.com/DBFocus/p/1827192.html
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