How to generate a CREATE TABLE statement for a given table in SQL Server

Shiji Pan

11 years ago

declare @table varchar(100)
set @table = 'my_table' -- set table name here
declare @sql table(s varchar(1000), id int identity)

-- create statement
insert into  @sql(s) values ('create table [' + @table + '] (')

-- column list
insert into @sql(s)
select
    '  ['+column_name+'] ' +
    data_type + coalesce('('+cast(character_maximum_length as varchar)+')','') + ' ' +
    case when exists (
        select id from syscolumns
        where object_name(id)=@table
        and name=column_name
        and columnproperty(id,name,'IsIdentity') = 1
    ) then
        'IDENTITY(' +
        cast(ident_seed(@table) as varchar) + ',' +
        cast(ident_incr(@table) as varchar) + ')'
    else ''
    end + ' ' +
    ( case when IS_NULLABLE = 'No' then 'NOT ' else '' end ) + 'NULL ' +
    coalesce('DEFAULT '+COLUMN_DEFAULT,'') + ','

 from information_schema.columns where table_name = @table
 order by ordinal_position

-- primary key
declare @pkname varchar(100)
select @pkname = constraint_name from information_schema.table_constraints
where table_name = @table and constraint_type='PRIMARY KEY'

if ( @pkname is not null ) begin
    insert into @sql(s) values('  PRIMARY KEY (')
    insert into @sql(s)
    	select '   ['+COLUMN_NAME+'],' from information_schema.key_column_usage
    	where constraint_name = @pkname
    	order by ordinal_position
    -- remove trailing comma
    update @sql set s=left(s,len(s)-1) where id=@@identity
    insert into @sql(s) values ('  )')
end
else begin
    -- remove trailing comma
    update @sql set s=left(s,len(s)-1) where id=@@identity
end

-- closing bracket
insert into @sql(s) values( ')' )

-- result!
select s from @sql order by id

From: http://stackoverflow.com/questions/21547/in-sql-server-how-do-i-generate-a-create-table-statement-for-a-given-table