Transact‐SQLでFizzBuzz問題を解いてみた

Posted on 2011年12月21日, 7:00 PM, by kimikato, under プログラミング.

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この記事の所要時間：約 0分46秒

haruiruka氏が

とつぶやいていたので、気分転換にTransact-SQLで書いてみました。

最初は簡単にwhileでループ処理して出力する方法。

declare @num int;
declare @max int;
set @num = 1;
set @max = 100;

while @num <= @max
	begin
		print
			case
				when @num % 3 = 0 and @num % 5 = 0 then 'FizzBuzz'
				when @num % 3 = 0 then 'Fizz'
				when @num % 5 = 0 then 'Buzz'	
				else cast(@num as nvarchar(5))
			end;

		set @num = @num + 1;
	end

declare @num int;

declare @max int;

set @num = 1;

set @max = 100;

while @num <= @max

begin

case

when @num % 3 = 0 and @num % 5 = 0 then 'FizzBuzz'

when @num % 3 = 0 then 'Fizz'

when @num % 5 = 0 then 'Buzz'

else cast(@num as nvarchar(5))

end;

set @num = @num + 1;

end

次に、ムダにtable型変数を定義し、select句で結果を出力。
ただただtable型変数を使ってみたかっただけです。

declare @num int;
declare @max int;
set @num = 1;
set @max = 100;

declare @temp table (
	id int not null
);

while @num <= @max
	begin
		insert into @temp values (@num);
		set @num = @num + 1;
	end

select
	case
		when id % 3 = 0 and id % 5 = 0 then 'FizzBuzz'
		when id % 3 = 0 then 'Fizz'
		when id % 5 = 0 then 'Buzz'
		else cast(id as nvarchar(5))
	end as result
from @temp

declare @num int;

declare @max int;

set @num = 1;

set @max = 100;

declare @temp table (

id int not null

);

while @num <= @max

begin

insert into @temp values (@num);

set @num = @num + 1;

end

select

case

when id % 3 = 0 and id % 5 = 0 then 'FizzBuzz'

when id % 3 = 0 then 'Fizz'

when id % 5 = 0 then 'Buzz'

else cast(id as nvarchar(5))

end as result

from @temp

もうちょっとスマートにする方法が浮かばず、なにか不完全燃焼チックです…
特にcase式で出力してる点が…

タグ: FizzBuzz問題, SQL Server, T-SQL, Transact-SQL
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