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Php versions – Why does my PHP ternary operator return "three" instead of "two" in versions < 5.3?

I’m having trouble understanding why my PHP code returns "three" instead of "two". Here’s the code:

php
<?php $a = 2; echo $a == 1 ? "one" : $a == 2 ? "two" : $a == 3 ? "three" : "others"; ?>

When I run this in PHP versions earlier than 5.3, the output is "three". Can someone explain why this happens and how the code works?

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2

Answers


  1. 0

    Your code is quite unreadable, it hurts my brain. Make it readable and you probably also solve any problems you have.

    How about this:

    
$number = 2;

switch ($number) {
   case 1 : echo "one"; break;
   case 2 : echo "two"; break;
   case 3 : echo "three"; break;
   default : echo "others"; break;
}

    This code is readable and, may I say, slightly more performant.

    There are also many other ways of doing it. Anything is better than multiple ternary operators on one line.

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  2. 0

    The ternary in PHP works in other direction compared to other programming languages.

    echo $a == 1 ? "one" : $a == 2 ? "two" : $a == 3 ? "three" : "others";

    is the same as

    echo (( $a == 1 ? "one" : $a == 2 ) ? "two" : $a == 3 ) ? "three" : "others";

    This can be reduced to CONDITION ? "three" : "others" where CONDITION is (( $a == 1 ? "one" : $a == 2 ) ? "two" : $a == 3 ). And CONDITION is always true.

    Compared to other programing languages, I think you want

    echo $a == 1 ? "one" :( $a == 2 ? "two" :( $a == 3 ? "three" : "others"));
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