Note that you rarely change a result set, so there is very little point in
returning a reference to it given PHP's shallow copy implementation. And
in your case you are just returning a resource id which you definitely
aren't going to change.
But if for some reason you feel it is important (please explain why) then
this is the syntax:
function & bdConsultar($strCon) {
$idRes = @mysql_query($strCon);
return $idRes;
}
$myID = & bdConsultar("Select * from users");
But again, I don't think you are quite understanding references.
-Rasmus
On Thu, 4 Apr 2002, javier wrote:
> I trying to code a kind of DB wrapper.
> So when is dbQuery turn I run into trouble.
> I read in php manual that refrences are not like C pointers. They just
> point to the same content.
>
> I want to return the result from a mysql_query
> but if I do something like this:
>
> function bdConsultar($strCon) {
> $idRes = @mysql_query($strCon);
> return $idRes;
> }
>
> Should be called like this?
>
> $myID = &bdConsultar("Select * from users");
>
>
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