Here's an idea -- patience is a virtue.  Some of us read the list when
we get a chance, and can't just jump on every e-mail because we
actually do work at work.

mysql> create table Orders (partno int unsigned not null, quant
tinyint unsigned not null);
Query OK, 0 rows affected (0.07 sec)

mysql> insert into Orders
VALUES(1254,5),(1414,5),(14758,1),(1254,6),(1024,3),(1254,1);
Query OK, 6 rows affected (0.00 sec)
Records: 6  Duplicates: 0  Warnings: 0

mysql> select partno,SUM(quant) from Orders where partno=1254 group by partno;
+--------+------------+
| partno | SUM(quant) |
+--------+------------+
| 1254   | 12         |
+--------+------------+
1 row in set (0.00 sec)

mysql> select repeat(concat(partno,' '),SUM(quant)) from Orders where
partno=1254 group by partno;
+--------------------------------------------------------------+
| repeat(concat(partno,' '),SUM(quant))                        |
+--------------------------------------------------------------+
| 1254 1254 1254 1254 1254 1254 1254 1254 1254 1254 1254 1254  |
+--------------------------------------------------------------+
1 row in set (0.00 sec)

So something like that.  I don't know that you can create more than
one record, though.

-Sheeri

On 4/7/06, Ed Reed <[EMAIL PROTECTED]> wrote:
> Anyone have an idea on this?
>
> Can anyone explain how I might be able to return a numbers of records based 
> on the sum of a quantity in a field in the same table? (After I read that it 
> sounds even confusing to me).
>
> Let me explain. I have records like this,
>
> Part#                Qty
> 1254                  5
> 1414                  2
> 14758                1
> 1254                  6
> 1024                  3
> 1254                  1
>
>
> Now if I did a query like this
> Select Part#, Sum(Qty) From table1 Group By Part# Where Part#=1254
>
> I would expect my results to look like this
> Part#           Sum(Qty)
> 1254               12
>
> But what I really want is this
> Part#
> 1254
> 1254
> 1254
> 1254
> 1254
> 12541254
> 1254
> 12541254
> 1254
> 1254
>
> So 12 virtual records for the count of the records returned from the Sum()
>
> Can someone help me with this?
>
> - Thanks
>
>
>
>
>
>
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>

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