[RESOLVED] Return False - error

LarryK

I'm trying to see why this code is returning a null rather than a zero. Here is the script:

<!DOCTYPE html PUBLIC
"-//W3C//DTD XHTML 1.0 Strict//EN"
"http://www.w3.org/TR/xhtmll/DTD/xhtmll-strict.dtd">

<head>

<title>Return Value</title>

</head>

<body>

<?php

$ReturnValue = 2 == 3 ;

echo "Number 1-$ReturnValue ";

$ReturnValue = "2" + "3" ;

echo "Number 2- $ReturnValue ";

$ReturnValue = 2 >= 3 ;

echo "Number 3-$ReturnValue ";

$ReturnValue = 2 <= 3 ;

echo "Number 4-$ReturnValue ";

$ReturnValue = 2 + 3 ;

echo "Number 5-$ReturnValue ";

$ReturnValue = (2 >= 3) && (2 > 3);

echo "Number 7-$ReturnValue ";

</body>

</html>

When I run it and expect to see -0 I see nothing or -

is this wrong?

Thanks for looking.

Larry

NogDog

// cast boolean to integer type:
$ReturnValue = (int) ((2 >= 3) && (2 > 3));
echo "Number 7-$ReturnValue<br />";

LarryK

thanks - that does the trick, but I'm wondering why the 4 returns a one as it was?

Installer

1 == true

Try this:

$exps = array(1 => '2 == 3', '"2" + "3"', '2 >= 3', '2 <= 3', '2 + 3 ', 7 => '(2 >= 3) && (2 > 3)');
foreach ($exps as $key => $exp) {
    $str = '<I>' . $key . '</I>}  ' . $exp . ' = ';
    $ret = eval('$ReturnValue = ' . $exp . ';');
    if (is_null($ReturnValue)) {
        $str .= 'null';
    } elseif ($ReturnValue === true) { 
        $str .= 'true';
    } elseif ($ReturnValue === false) { 
        $str .= 'false';
    } elseif (is_string($ReturnValue)) { 
        $str .= '"' . $ReturnValue . '"';
    } else { 
        $str .= $ReturnValue;
    }
    echo $str . '<br />';
}

LarryK

Thanks - that sheds further light on this.