PHP : How to echo return array method?

Question

I have a class that has this method:

public function user_status()
    {    
            $json = //some code here...
            $result = json_decode($json, true);
            $status = array(
                'key1' => $val1,
                'key2' => $val2
            );
            return $status;

    }

when calling like this:

$user = new Class();
echo $user->user_status();

it just returns the word Array but when dumping like so:

echo "<pre>User: ", print_r($user->user_status(), TRUE), "</pre>";

it prints the array as expected:

User: Array
(
    [key1] => val1
    [key2] => val2
)

I want to output the array values so I tried calling it like below but its giving me a parse error.

echo $user->user_status()->['key2'];

Whats the proper syntax to output the array values?

John Conde · Accepted Answer · 2014-01-27 17:58:14Z

7

In PHP 5.3 or older:

$array = $user->user_status();
echo $array['key2'];

In PHP 5.4 and newer:

echo $user->user_status()['key2'];

PHP 5.4 introduced array dereferencing which allows us to shorten the syntax a bit.

answered Jan 27, 2014 at 17:58

John Conde

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4 Comments

Liam Sorsby Over a year ago

You have to echos! not sure why this has been down voted though

LohanHill1 Over a year ago

I tried the latter approach but its giving me a parse error. must be the php version.

Amal Over a year ago

+1 - I like this answer best. Clear and concise. I would add 5.4+ stuff to the top though :)

LohanHill1 Over a year ago

I like @zavg answer just because the variable naming is more clearer but I will choose this answer for completeness! thanks all for your help.

efux · Accepted Answer · 2014-01-27 17:58:21Z

3

Try

echo $user->user_status()['key2'];

The return value of user_status() itself is an array, so you can "treat the function call as if it would be the name of a variable".

answered Jan 27, 2014 at 17:58

efux

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1 Comment

LohanHill1 Over a year ago

yeah. tried that before but its giving me a parse error. I suspect its because my php version does not support it. but thanks anyway!

zavg · Accepted Answer · 2014-01-27 17:58:23Z

3

$status = $user->user_status();
echo $status['key2'];

answered Jan 27, 2014 at 17:58

zavg

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Comments

php-dev · Accepted Answer · 2014-01-27 17:58:08Z

2

This is the syntax

$array = $user->user_status();

echo $array['key2'];

answered Jan 27, 2014 at 17:58

php-dev

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