How I can get the real data type result instead of integer data type?

On Mon, 2006-03-20 at 21:10, Wei Wei wrote:

In a query, there is something like

order by count(id)/age

where both id and age are the integer data type.

From a query result, I believe the operation count(id)/age yields a integer. I need it in real data type. After searching the online document, I haven't found any related information. Can someone help me out on this problem, please.

Thanks,

w

Try:

order by count(id)/age::float

regards,
Sig.Gunn

On Mon, Mar 20, 2006 at 13:10:51 -0800,
Wei Wei <wei725@lycos.com> wrote:

In a query, there is something like

order by count(id)/age

where both id and age are the integer data type.

From a query result, I believe the operation count(id)/age yields a integer. I need it in real data type. After searching the online document, I haven't found any related information. Can someone help me out on this problem, please.

You can cast the expressions. Something like:
order by count(id)::float/age::float

"Sigurdur Gunnlaugsson" <sig@fjolnet.net>

On Mon, 2006-03-20 at 21:10, Wei Wei wrote:

Try:

order by count(id)/age::float

Or you can use the standard grammer:

order by cast(count(id)/age as float)

Regards,
William ZHANG

On Tue, Mar 21, 2006 at 21:16:03 +0800,
William ZHANG <uniware@zedware.org> wrote:

"Sigurdur Gunnlaugsson" <sig@fjolnet.net>

On Mon, 2006-03-20 at 21:10, Wei Wei wrote:

Try:

order by count(id)/age::float

Or you can use the standard grammer:

order by cast(count(id)/age as float)

Don't you have to cast before the divide? I think the above will convert
the truncated result to float.
However he could do something like:
order by count(id)/cast(age as float)

Bruno Wolff III <bruno@wolff.to> writes:

Try:
order by count(id)/age::float

Or you can use the standard grammer:

order by cast(count(id)/age as float)

Don't you have to cast before the divide?

Yeah. The :: case is OK because :: binds more tightly than /
but the second suggestion is wrong :-(

regards, tom lane