if calculation help...

[fikse]

can someone explain why this code gives a parse error?

<?

$caryear = '1996';
$notifycaryearop = '>=';
$notifycaryear = '2000';


if ($caryear $notifycaryearop $notifycaryear) 

{echo "yes";}

else

{echo "no";}



?>

[laserlight]

Because you're trying to store a string literal into a variable, then use that variable as an operator, rather than using the string literal as the symbol for the operator.

The crux of it is that:
$caryear $notifycaryearop $notifycaryear
is simply wrong syntax.

Even if it were accepted, it would become:
'1996 >= 2000'
not
1996 >= 2000

PHP strings can be interpreted as integers, but the same doesnt apply where operators instead of numbers are concerned.

[sam159]

use someting like this

        function compare($statement1,$statement2,$sign)
        {
        if ($sign == "==")
        	{
            if ($statement1 == $statement2) {return TRUE;}
        	}
        if ($sign == "===")
        	{
            if ($statement1 === $statement2) {return TRUE;}
        	}
        if ($sign == ">")
        	{
            if ($statement1 > $statement2) {return TRUE;}
        	}
        if ($sign == ">=")
        	{
            if ($statement1 >= $statement2) {return TRUE;}
        	}
        if ($sign == "<")
        	{
            if ($statement1 < $statement2) {return TRUE;}
        	}
        if ($sign == "<=")
        	{
            if ($statement1 <= $statement2) {return TRUE;}
        	}
        if ($sign == "!=")
        	{
            if ($statement1 != $statement2) {return TRUE;}
        	}
        if ($sign == "!==")
        	{
            if ($statement1 !== $statement2) {return TRUE;}
        	}
        if ($sign == "<>")
        	{
            if ($statement1 <> $statement2) {return TRUE;}
        	}
        }
        $sign = "<>";
        $number1 = "1";
        $number2 = "2";
        if (compare($number1,$number2,$sign)==TRUE)
        {echo "$number1 $sign $number2";}