Re: [PHP] Is this possible ?

[Dagfinn Reiersøl]

An object-oriented way of doing it: keep the extra result in a member variable and
get it separately. You might not find this necessary, but it gets more useful
in more complex cases.
class MyClass {
 var $num;
 Function MyFunc(){
   if(isset($_POST['var'])){
     $sql = mysql_query("select * from table_name where field=\"$_POST[var]\"
         ");
     $this->num = mysql_num_rows($sql); 

       $returnsomething ="blah blah";
   }
   return $returnsomething;
 }
 function getCount() { return $this->num; }

}



Dave Carrera wrote:

Hi List,

I have a function that makes a call to mysql based on certain vars.

---example----

Function MyFunc(){
if(isset($_POST[var])){
$sql = mysql_query("select * from table_name where field=\"$_POST[var]\"
");
$returnsomething ="blah blah";
}
return $returnsomething;
}
And that all works fine no probs here but.....

I want to use a result somewhere in my script that is not returned by
return. Let me show you...
---example----

Function MyFunc(){
if(isset($_POST[var])){
$sql = mysql_query("select * from table_name where field=\"$_POST[var]\"
");
$num = mysql_num_rows($sql); // I want to use this result outside this
function.
$returnsomething ="blah blah";
}
return $returnsomething;
}
So $num contains a number that I want to use outside the function which is
not covered by return.
I know return stops a script and returns what I want it to return but how do
I send out of the function the var I want.
I have tried $GLOBAL[var]=$num; but that don’t work, but I thought I
would'nt anyway just tried it and yes I know I have to declare it inside my
new function using global $var; to use it.
So I ask is this achiveable or how can I do this.

Thank you in advance

Dave C

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