Why ternary operation return false result?

Question

I have custom ternary operation where I get user gender from numbers:

Code:

$value = 0;
$gender = $value === 0 ? 'Male' : $value === 1 ? 'Female' : 'Other';
echo $gender;

In my think this code must return Male but return Female. Here is example.

Or my ternary operation incorrect for getting Male result?

My ternary operation not equal with this code?

Code:

if($value === 0) $gender = 'Male';
elseif($value === 1) $gender = 'Female';
else $gender = 'Other';

Dont know why, but you should add parantheses for else-if ternary like this: `$value === 0 ? 'Male' : ($value === 1 ? 'Female' : 'Other');` — Code Spirit, Sep 05 '19 at 08:26

score 4 · Answer 1 · answered Sep 05 '19 at 08:26

4

It's a precedence issue, your code is evaluated as:

$gender = ($value === 0 ? 'Male' : $value === 1) ? 'Female' : 'Other';

but what you want is:

$gender = $value === 0 ? 'Male' : ($value === 1 ? 'Female' : 'Other');

This is because PHP evaluates the ternary operator from left to right. See the manual.

answered Sep 05 '19 at 08:26

Nick

There're tons of duplicates, yet you decided toi create anoter one. – u_mulder Sep 05 '19 at 08:28
@u_mulder I tried googling and SO search but can't find it. Can you close it? – Nick Sep 05 '19 at 08:29
1

@u_mulder thanks. I have added it to my dupes list. My problem was I only searched for "not working", which brings up a lot of questions which are not the same as this. – Nick Sep 05 '19 at 08:32

score 0 · Answer 2 · answered Sep 05 '19 at 08:27

0

You need to add some brackets, take reference from here

<?php

$value = 0;
$gender = $value === 0 ? 'Male' : ($value === 1 ?'Female' : ('Other'));
echo $gender;

Check Example: here

answered Sep 05 '19 at 08:27

Ankur Tiwari

2 Answers2