Is this if and this ternary equivalent?

(($no%3==1)? $num = 1: ($no%3==2))? $num = 2: ($no%3==0)) $num =3:false;
And
if ($no 1%3==1) {
$num 1=1;
}elseif ($no 1%3==2) {
$num 1=2;
}elseif ($no 1%3==0) {
$num 1=3;
}
Tried, if the display, and the ternary does not show, do not know where the wrong

Reply to discussion (solution)

$num =3:false; This place is in front of the assignment, the back is directly a false.

Look at the battle, the landlord is the SHEEP_APP3 bar

$map = Array (3, 1, 2), $num 1 = $map [$no% 3];

($no%3==1)? $num = 1: (($no%3==2)? $num = 2: (($no%3==0) $num =3:false));

Your ternary and if are not equivalent, take a closer look at your ternary will only produce two results, $num = 3 or False

The above answer is not to the landlord, to only use if, switch's pot friends to provide another way of thinking, is to use Hashtable, branch for a long while its efficiency is the highest

