PHP code meaning and usage

Question

What does this piece of PHP actually mean and how could it be displayed differently?

'S_DISPLAY_VEHICLE_OWNER'   => ($owned == 'MODERATE' || $owned == 'YES') ? 1 : 0

See also: http://stackoverflow.com/q/2153180/1935077 – Petr R. Jul 16 '13 at 09:59 — Petr R., Jul 16 '13 at 09:59

score 5 · Accepted Answer · answered Jul 16 '13 at 09:57

'S_DISPLAY_VEHICLE_OWNER'   => ($owned == 'MODERATE' || $owned == 'YES') ? 1 : 0

another look:

if ($owned == 'MODERATE' || $owned == 'YES'){
 $result = 1;
}else{
 $result = 0;
}

'S_DISPLAY_VEHICLE_OWNER'   => $result

score 2 · Answer 2 · answered Jul 16 '13 at 09:58

2

It sets S_DISPLAY_VEHICLE_OWNER with a shorthand notation of if/else.

if($owned == 'MODERATE' OR $owned == 'YES') {

    return 1

}else{

    return 0

}

answered Jul 16 '13 at 09:58

Pieter

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score 0 · Answer 3 · answered Jul 16 '13 at 09:57

0

if $owned equals 'MODERATE' or 'YES' then the S_DISPLAY_VEHICLE_OWNER variable will be populated with 1 else 0

answered Jul 16 '13 at 09:57

Nil'z

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score 0 · Answer 4 · answered Jul 16 '13 at 09:57

0

Basically in plaintext the expression ($owned == 'MODERATE' || $owned == 'YES') ? 1 : 0 means

if owned is moderate or owned is yes then 1 else 0

The field S_DISPLAY_VEHICLE_OWNER in your array is set to 1 or 0 depending on the value of $owned.

answered Jul 16 '13 at 09:57

apartridge

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Expedito · Answer 5 · 2013-07-16T11:08:50.747

0

It's assigning a value of 1 or 0 to an associative array if the value of $owned is either "MODERATE" or "YES." So, if the array were $arr, then it could be rewritten as:

if ($owned == 'MODERATE' || $owned == 'YES'){
   $arr['S_DISPLAY_VEHICLE_OWNER'] = 1;
} else {
   $arr['S_DISPLAY_VEHICLE_OWNER'] = 0;
}

edited Jul 16 '13 at 11:08

answered Jul 16 '13 at 09:59

Expedito

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PHP code meaning and usage

5 Answers5