If else with Ternary operator

Question

Below is the example:
If I write like:

<?php
$var = 'A';
echo    ($var == 'B' || $var == 'C')  ? 'B or C' : 'A';
?>

//Out will be "A" But if I write like below:

<?php
$var = 'A';
echo    ($var == ('B' || 'C'))  ? 'B or C' : 'A';
?>

It give me out put as "B or C".
Here ($var == ('B' || 'C')) is not correct or i am missing something?

Created PHP fiddle: http://phpfiddle.org/main/code/wju-46r

Because in the second you're doing an OR of 'B' and 'C', then comparing the result of that with $var; 'B' OR 'C' will evaluate to a Boolean (true), so you get type casting $var to Boolean to compare with Boolean true — Mark Baker, Feb 13 '14 at 07:55

score 1 · Accepted Answer · answered Feb 13 '14 at 07:56

This has no relation to ternary operator. It's about type juggling and comparison.

In second case, you're doing 'B' || 'C' which will be treated as true - since || is logical operator. So 'A' == true is true because of type juggling, thus B or C will be your result

score 0 · Answer 2 · answered Feb 13 '14 at 08:01

0

echo ($var === ('B' || 'C')) ? 'B or C' : 'A';

This is the right answer you must have to check data type also.

answered Feb 13 '14 at 08:01

Diwakar upadhyay

434
5
14

1

this will not work if you give $var = 'B'... it will give you 'A' – Kalpit Feb 13 '14 at 08:05

If else with Ternary operator

2 Answers2