combining multiple rows into one row

Hello Everyone,

This seems simple, but I am finding it hard to do it without a temp table or
a cursor. I need to do it without both.

The values of PRTHASH below need to become a single row

select TOP(5) CUST_ID,
hashbytes('SHA1', CAST(CUST_ID AS VARCHAR) +
',' + NAME1 +
',' + CAST(CUST_TIMESTAMP AS VARCHAR) )
AS 'PRTHASH'
from cust
where Cust_ID > @start
order by Cust_ID

I can do the following if I store CUST_ID and PRTHASH in a TEMP table

select @part ='';
select @part= @part + cast(PRTHASH AS VARCHAR)+ ',' from TEMP

But without a temp table or cursor, I am lost. I believe a correlated
subquery or replace statements with CASE can be used, but can't figure out
how. Even if I do use them, wouldn't this require writing many select
statements, if I select TOP 100 rows.

Please give me some hint if there exists one.

I have seen some posts below that answer a similar question, but they either
use a function or save a table somewhere. I am not supposed to do that. I
also need to use a command that most major databases will understand. The
single row strings must fall out of select statements. Is that possible?
Thanks for your help!




"mayub" wrote:

> Hello Everyone,
>
> This seems simple, but I am finding it hard to do it without a temp table or
> a cursor. I need to do it without both.
>
> The values of PRTHASH below need to become a single row
>
> select TOP(5) CUST_ID,
> hashbytes('SHA1', CAST(CUST_ID AS VARCHAR) +
> ',' + NAME1 +
> ',' + CAST(CUST_TIMESTAMP AS VARCHAR) )
> AS 'PRTHASH'
> from cust
> where Cust_ID > @start
> order by Cust_ID
>
> I can do the following if I store CUST_ID and PRTHASH in a TEMP table
>
> select @part ='';
> select @part= @part + cast(PRTHASH AS VARCHAR)+ ',' from TEMP
>
> But without a temp table or cursor, I am lost. I believe a correlated
> subquery or replace statements with CASE can be used, but can't figure out
> how. Even if I do use them, wouldn't this require writing many select
> statements, if I select TOP 100 rows.
>
> Please give me some hint if there exists one.
>
>
>