PHP Using IF Variables

$a = "1";
$b = "2";
if ($a == $b) {
is true
} else {
is false
}

The above code works just fine.

$filter = "$a == $b";
if ($filter) {
is true
} else {
is false
}
The above code is always true.

I am generating the $filter from user input. I do not know in advance if the operator will be "==" or "!=" . Does anyone know a solution?

**** Bowers

Sorry if this is here twice . . . I hit tab-enter and inadvertently submitted. Here's my complete response . . .

Your if statement is checking only that the variable $filter has a value. That's why it's always true. If your user only has two options (== or !=) set each of these possibilities to a value, check for that value, and use a switch. For example:

assuming you're getting these values for each possibility,
== sets $filter = "equal"
!= sets $filter = "notEqual"

switch ($filter) {

case "equal":
equal code here . . . $a == $b;
break;

case "notEqual":
not equal code here . . .$a != $b;
break;
}

I thought that might be the case but i wanted to be sure. It is not the == != that is the problem. There can be up to 20 different combos. I will use SQL to solve the problem.

Thanks for your response.

**** Bowers