Finding value of smaller value and swap in php

Question

$ var1 = 22; $ var2 = 10; echo $ var1 = ($ var1 <$ var2)? $ var1: $ var2; // smaller var echo '
'; echo $ var2 = ($ var1> $ var2)? $ var1: $ var2; // more than var

I expect it to print 10 and 22, but print 10 and 10. Any ideas what I am doing wrong?

thank

UPDATE Thanks to everyone.

    $min = min($var1, $var2); 
    $max = max($var1, $var2);

    $var1 = $min;
    $var2 = $max;

Answer 1

@unicornaddict already decided your problem, but for simplicity, you can use minand maxthe PHP function.

echo min($var1, $var2), '<br/>', max($var1, $var2);

Answer 2

You reassign variables to echo.

// $var1 is being assigned minimum of 10,22 which is 10.
// after this $var1 and $var2 will both be 10.
echo $var1 = ($var1 < $var2) ? $var1 : $var2; 

:

echo ($var1 < $var2) ? $var1 : $var2; // prints min.
echo '<br />';
echo ($var1 > $var2) ? $var1 : $var2; // prints max.

EDIT:

$var1, :

if($var1 > $var2) { // if $var1 is larger...swap.
        list($var1,$var2)  = array($var2,$var1);
}

Answer 3

$var1. , 10 > 10.

$var1 = 22;
$var2 = 10;
echo $var1 = (10 < 22) ? 22 : 10; //smaller var -> $var1 now has the value 10
echo '<br />';
echo $var2 = (10 > 10) ? 22 : 10; //greater var -> 10 is not greater than 10, so $var2 gets a value of 10.

Answer 4

10 $var1 , 10.

Answer 5

echo $var1 = ($var1 < $var2) ? $var1 : $var2; //smaller var

10 $var1. 10. , ?

Answer 6

. min,

echo min($var1, $var2);

Answer 7

, , , . , , , $var1 $var2 .

$var1 = 22;
$var2 = 10;
if ($var1 > $var2) {
    list($var1, $var2) = array($var2, $var1);
}
// $var1 will now be smaller than (or equal to!) $var2