Question

我使用以下代码在SQL Server中创建序列。但它将错误显示为未知对象类型。请给出解决方案

这是我的代码：

create sequence seqval start with 100 increment by 1 minvalue 0 maxvalue 0 no cycle  
 no cache;

提前致谢

Answer 1

你可以这样做。

--Create a dummy TABLE to generate a SEQUENCE. No actual records will be stored.
CREATE TABLE SequenceTABLE
(
    ID BIGINT IDENTITY  
);
GO

--This procedure is for convenience in retrieving a sequence.
CREATE PROCEDURE dbo.GetSEQUENCE ( @value BIGINT OUTPUT)
AS
    --Act like we are INSERTing a row to increment the IDENTITY
    BEGIN TRANSACTION;
    INSERT SequenceTABLE WITH (TABLOCKX) DEFAULT VALUES;
    ROLLBACK TRANSACTION;
    --Return the latest IDENTITY value.
    SELECT @value = SCOPE_IDENTITY();
GO

--Example execution
DECLARE @value BIGINT;
EXECUTE dbo.GetSEQUENCE @value OUTPUT;
SELECT @value AS [@value];
GO

Answer 2

创建一个Numbers表;这是关于这个主题的SO问题。我们称之为dbo.Number。

拥有一个包含标识列的表。将种子和步骤设置为适当的值：

create table dbo.SequenceGenerator(ID int identity(1, 1), dummy int);

然后从数字表中插入值并捕获新生成的标识值：

declare @HowMany int = 3;  -- This determines how large a sequence you receive
                           -- at each itteration
declare @NewSequenceValue table (ID int);

insert dbo.SequenceGenerator(dummy)
output INSERTED.ID 
    into @NewSequenceValue
select Number from dbo.Numbers
where Number <= @HowMany;

select * from @NewSequenceValue;

请务必不时DELETE .. dbo.SequenceGenerator，否则它会变大，无其他价值。请勿TRUNCATE - 将IDENTITY列重置为其初始声明的种子值。

Answer 3

SQL Server 2008无法创建序列，Sequence对象通过当前版本应用于SQL Server 2012.

https://msdn.microsoft.com/es-es/library/ff878091(v=sql.120).aspx

您可以在表格中使用IDENTITY，例如：

CREATE TABLE Person(
    Id int IDENTITY(1,1) NOT NULL PRIMARY KEY,
    Name varchar(255) NOT NULL
);

IDENTITY的起始值为1，每个新记录的起始值都会增加1。

http://www.w3schools.com/sql/sql_autoincrement.asp

Answer 4

WITH N0 as (SELECT 1 as n UNION ALL SELECT 1)
,N1 as (SELECT 1 as n FROM N0 t1, N0 t2)
,N2 as (SELECT 1 as n FROM N1 t1, N1 t2)
,N3 as (SELECT 1 as n FROM N2 t1, N2 t2)
,nums as (SELECT ROW_NUMBER() OVER (ORDER BY (SELECT 1)) as num FROM N3)
SELECT * FROM nums

Answer 5

在SQL Server 2008中，我们无法轻松使用Sequence。

您可以使用CTE（公用表表达式）在SQL Server 2008中生成序列

WITH NUM_GEN (n) AS
     ( 
            SELECT 1 
            UNION 
                  ALLSELECT n+1 
            FROM  NUM_GEN 
            WHERE n+1< MAX_VALUE 
     ) 
SELECT n 
FROM   NUM_GEN

Answer 6

你确定你正在运行2012吗？我没有遇到麻烦：

CREATE SEQUENCE seqval
START WITH 100
INCREMENT BY 1 
minvalue 100 maxvalue 10000 NO CYCLE

您的0,0值为我生成了一个语法错误，但是一个清晰而简单的错误。

The minimum value for sequence object 'seqval' must be less than its maximum value.

如何在SQL Server 2008中创建序列

6 个答案: