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  1. declare @cnt int,@ sql varchar(1000) ---計算筆數 select @cnt = count(*) from table1 set @cnt = @cnt / 3 -----------------歸零 update table1 set 完成期限 = '00000000' --組 SQL 語法 ----------------------------1.20090410 set @ sql = 'update table1 set 完成期限 = '...

  2. DECLARE @i INTEGER DECLARE @sum INTEGER SELECT @i = 1...

  3. declare @n int,@nn int set @nn=0 set @n=1 while @n<=1000 begin set @nn=@nn+@n set @n=@n+1 end select @nn (這就是答案) 2009-08-11 17:39:08 補充: declare @上底 int,@下底 int,@高 int set @上底=1 set @下底=1000 set @高=1000 select...

  4. ...補充. ps: 以上是 T- SQL (MS SQL )的語法. 2010-03-29 23:07:54 補充: "我的目的是...還是 varchar? 解法稍有不同.. 2010-03-30 09:21:34 補充: declare @YYYY char(4) declare @MM char(2) declare @...

  5. DECLARE @C int SET @C = 100 SELECT (9.0*@C)/5.0...

  6. 如果要再 SQL 裡做的話我是會寫 PROCEDURE 利用CURSOR..., sysname, ProcedureName> AS DECLARE @type char(30),@name char(30),@pulse char(50),@describe char(50),@prescription...

  7. ... sp_AppendToFile1(@FileName varchar(255)) AS DECLARE @FS int, @OLEResult int, @FileID...

  8. declare @X int,@Y int,@i int,@total1 int,@total2 int set @X=50 set @...amp;#39;(1+2+3+4+...+X)+1+2+3+4+...+Y =&#39;+ltrim(str(@total2)) declare @limit int,@i int,@total int set @limit=50000 set @i=1 set @total=0 while @total+@i...

  9. declare @i smallint,@j smallint,@str varchar(100) set @i=1 while @i<=9 begin select @j=1,@str= while @j<=@i select @str=@str +cast(@j as char(1)) + * +cast(@i as char(1)) + = +cast(@i*@j as char(2)) +space(2) ,@j=@j+1 print @str set @i=@i+1 end

  10. ...大哥 容小弟放肆一下 Sorry ============= SQL 語法是可以做到的,但提問大大似乎轉移了問題...答案--最小值 Function 內容如下: DECLARE @RLT decimal(10,2) 1.利用 資料指標...