打印连续的数字

Question

我如何使用单个 sql 查询打印连续的数字..如 1 , 2 , 3 ...

Answer 1

;with cte
as 
(
select 1 [sequence]
union all
select [sequence]+1 from cte where [sequence]<100
)
select * from cte

尝试，通过使用通用类型表达式我们可以做到......它的工作原理

Answer 2

不确定我是否正确理解了您的问题，但如果您只想PRINT连续的数字，我不明白为什么您不能执行以下操作：

DECLARE @a INT
SET @a = 1
WHILE @a <= 10
BEGIN
PRINT @a
SET @a += 1
END

Answer 3

你想获取sql结果集中的行号吗？尝试以下操作：

SET @line = 0; 
SELECT @line := @line + 1, some_field FROM table_name;

对于 SQL Server：http://msdn.microsoft.com/en -us/library/ms186734.aspx

SELECT FirstName, LastName, ROW_NUMBER() OVER(ORDER BY SalesYTD DESC) AS 'Row Number',     SalesYTD, PostalCode 

FROM Sales.vSalesPerson

WHERE TerritoryName IS NOT NULL AND SalesYTD <> 0;

Answer 4

像...

  select
            (a3.id + a2.id + a1.id + a0.id) as id
        FROM 
        /* create the tables to be used for the cartesian join */
        (
            select 0 id UNION ALL 
            select 1 UNION ALL
            select 2 UNION ALL
            select 3 UNION ALL
            select 4 UNION ALL
            select 5 UNION ALL
            select 6 UNION ALL
            select 7 UNION ALL
            select 8 UNION ALL
            select 9
        ) as a0,    
        (
            select 0 id UNION ALL
            select 10 UNION ALL
            select 20 UNION ALL
            select 30 UNION ALL
            select 40 UNION ALL
            select 50 UNION ALL
            select 60 UNION ALL
            select 70 UNION ALL
            select 80 UNION ALL
            select 90  
        ) as a1, 
        (
            select 0 id UNION ALL
            select 100 UNION ALL
            select 200 UNION ALL
            select 300 UNION ALL
            select 400 UNION ALL
            select 500 UNION ALL
            select 600 UNION ALL
            select 700 UNION ALL
            select 800 UNION ALL
            select 900
        ) as a2, 
        (
            select 0 id UNION ALL
            select 1000 UNION ALL
            select 2000 
        ) as a3 
    order by id asc ;

Answer 5

DECLARE @NUM INT, @COUNT INT , @NUM1 INT , @SPACE INT
SET @NUM=1 SET @COUNT=1000 SET @NUM1=0 

WHILE(@NUM<=@COUNT)

BEGIN

WHILE (@NUM1<=@NUM) 

BEGIN
DECLARE @STORE VARCHAR(MAX)
SET @NUM1=@NUM1+1 --1
SET @SPACE = (@COUNT-@NUM) --3

SET @STORE=ISNULL(@STORE,'')+SPACE(1)+CAST(@NUM1 AS VARCHAR(MAX))--1
PRINT (SPACE(@SPACE)+@STORE)

IF(@NUM<=@COUNT)
BEGIN
SET @NUM=@NUM+1
END
END


SET @NUM1=0

END

以三角形格式打印数字

Answer 6

declare @digits table (id int,value int )
insert @digits values (1,80),(1,90),(1,100),(1,200),(1,210),(1,9),(1,10),(1,12),
(2,8),(2,9),(2,11),(2,12),
(3,2),(3,4),(3,5),(3,7)

select distinct id,stuff(convert(varchar(max),
(
    select 
        ',' + case 
            when min(value) = max(value) then convert(varchar(10), min(value)) 
            else convert(varchar(10), min(value)) + '-' + convert(varchar(10), max(value)) 
        end
    from (select row_number() over (partition by id order by id,value) as seq, value,id from @digits) data
    where data.id = s.id
    group by (value/10) - seq
    for xml path('')
)), 1, 1, '') as result from @digits s

Answer 7

declare @digits table (id int,value int )
insert @digits values (1,1),(1,2),(1,3),(1,6),(1,8),(1,9),(1,10),(1,12),
(2,8),(2,9),(2,11),(2,12),
(3,2),(3,4),(3,5),(3,7)

select distinct id,stuff(convert(varchar(max),
(
    select 
        ',' + case 
            when min(value) = max(value) then convert(varchar(10), min(value)) 
            else convert(varchar(10), min(value)) + '-' + convert(varchar(10), max(value)) 
        end
    from (select row_number() over (partition by id order by id,value) as seq, value,id from @digits) data
    where data.id = s.id
    group by value - seq
    for xml path('')
)), 1, 1, '') as result from @digits s