恩...我也玩了一下@@..

declare @Tmp_Table1 as table(
	NAME nvarchar(50)
	,NUM int
)
insert into @Tmp_Table1
values('APPLE',1),('APPLE',2),('APPLE',4),('APPLE',6)
,('SAMSUNG',3),('SAMSUNG',5),('SAMSUNG',7),('SAMSUNG',10),('SAMSUNG',12),('SAMSUNG',13)

declare @Tmp_Table2 as table(
	NAME nvarchar(50)
	,Sub_Str nvarchar(Max)
)

declare @i int,@Max int,@NAME nvarchar(50),@Num_Str nvarchar(max)
declare @Sub_i int,@Sub_Max int
set @i = 1
set @Sub_i = 0
set @Num_Str = ''

select @Max = count(*)
from (
	select NAME
	from @Tmp_Table1 as a
	group by NAME
) as k

while(@i <= @Max)
begin
	
	select @Sub_Max = count(*)
	from (
		select Row_Number() Over(order by NAME) as Sort
		,NAME
		from (
			select NAME
			from @Tmp_Table1 as a
			group by NAME
		) as k
	) as k
	left join @Tmp_Table1 as b on k.NAME = b.NAME
	where Sort = @i

	select @NAME = k.NAME
	,@Num_Str = @Num_Str + Convert(varchar,b.NUM)
	+ (
		case when @Sub_Max = Row_Number() Over(order by k.NAME) 
		then ''
		else
			case when Row_Number() Over(order by k.NAME) % 2 = 0 then ',' else '-' end
		end
	)
	from (
		select Row_Number() Over(order by NAME) as Sort
		,NAME
		from (
			select NAME
			from @Tmp_Table1 as a
			group by NAME
		) as k
	) as k
	left join @Tmp_Table1 as b on k.NAME = b.NAME
	where Sort = @i

	insert into @Tmp_Table2
	values(@NAME,@Num_Str)
	
	set @Num_Str = ''

	set @i = @i + 1
end

select * from @Tmp_Table2

因為沒有給name跟另外一張表格的關聯結構
我這邊只做純num版本的

另外你的DB應該是SQL Server?

第一版 (SQL Fiddle連結):

備註:

• 沒有連續值情況要特別處理，舉例1、2、10
  10沒有連續，所以我把它當成另外一個表格額為處理，最後標示為1-2、10

with CTE as (
  select T1.num,T2.num n2,T3.num n3 from Table1 T1
  
  --以下兩個join目的是取得每組最小跟最大的連貫值
  left join Table1 T2 on T2.num = T1.num + 1
  left join Table1 T3 on T3.num = T1.num - 1
  
  --假如沒有null，代表是中間連貫值不需要考慮
  --假如T2.num跟T3.num都是null代表沒有連續值
  where T2.num is  null or T3.num is  null 
)
--有連續值，按數字大小排序
,CTE2 as (
    select num 
    ,ROW_NUMBER() OVER(ORDER BY num) rank
    from CTE 
    where not (n2 is null and n3 is null)
)
--連續組合-跟不連續union
,CTE3 as (
  --有連續值組合
  select convert(nvarchar(50),T1.num) + '-' + convert(nvarchar(50),T2.num) num
  --藉由mod2來區分組最小值，跟最大值 
  from (select * from CTE2 where rank % 2 = 1) T1
  left join (select * from CTE2 where rank % 2 = 0) T2 on T1.rank + 1 = T2.rank
  union all
  --沒有連續值
  select convert(nvarchar(50),num) from CTE
  where (n2 is null and n3 is null)
)
select STRING_AGG (num, ',') num from CTE3

結果:

主要邏輯寫在註解裡面

你先看一下，有不了解的在跟我說 :-)