PHP echo array with an array?

Question

I create an array which works fine:

$user_list_args = array(
 'role__in' => array('editor', 'author', 'administrator' ),
);
$user_list = get_users($user_list_args);
$user_list_array = array();
foreach ( $user_list as $user ) {
 $user_list_array[] = $user->user_nicename;
}

I then want to display that in an another array which is set up as follows:

array(
 'id' => 'task-users',
 'label' => 'User(s)',
 'type' => 'select',
 'options' => array(
  'tom',
  'jerry',
 ),
),

I want to change the options to be dynamic from my array as so:

private $fields = array(
 array(
  'id' => 'task-users',
  'label' => 'User(s)',
  'type' => 'select',
  'options' => array($user_list_array),
 ),
);

This generates an error:

Fatal error: Constant expression contains invalid operations

Can someone point me in the right direction? I am using PHP7 if that helps but should be usable on older versions too.

Thanks

Why are you creating the $user_list_array like that? If you want that array to contain the user's nice names, you only need to do $user_list_array[] = $user->user_nicename; — Parziphal
– Parziphal, Commented Dec 11, 2016 at 2:13
$newArray[] = print_r($oldArray,true); does this do what you need? — Mm-Art-In
– Mm-Art-In, Commented Dec 11, 2016 at 22:50

You Old Fool · Accepted Answer · 2016-12-11 23:05:01Z

1

You can't declare class properties using variables. If you need to assign values like that to the property use a constructor method and do it there. Maybe something like this:

private $fields;

public function __construct() {
    // your code to define $user_list_array somewhere here
    // or passed to constructor when you instantiate the class

    $this->fields = array(
        array(
            'id' => 'task-users',
            'label' => 'User(s)',
            'type' => 'select',
            'options' => array($user_list_array),
        )
    );
}

http://php.net/manual/en/language.oop5.properties.php:

This declaration may include an initialization, but this initialization must be a constant value--that is, it must be able to be evaluated at compile time and must not depend on run-time information in order to be evaluated.

edited Dec 11, 2016 at 23:05

answered Dec 11, 2016 at 22:52

You Old Fool

23.2k13 gold badges92 silver badges118 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

user557846 · Accepted Answer · 2016-12-11 02:10:55Z

0

your joining not echoing and this is completely valid:

<?php

$user_list_array=array(
  'tom',
  'jerry',
 );

$f= array(
 'id' => 'task-users',
 'label' => 'User(s)',
 'type' => 'select',
 'options' => array($user_list_array),
);

print_r($f);

output:

Array
(
    [id] => task-users
    [label] => User(s)
    [type] => select
    [options] => Array
        (
            [0] => Array
                (
                    [0] => tom
                    [1] => jerry
                )

        )

)

OR:

<?php

$user_list_array=array(
  'tom',
  'jerry',
 );

$f= array(
 'id' => 'task-users',
 'label' => 'User(s)',
 'type' => 'select',
 'options' => $user_list_array,
);

print_r($f);

output:

Array
(
    [id] => task-users
    [label] => User(s)
    [type] => select
    [options] => Array
        (
            [0] => tom
            [1] => jerry
        )

)

answered Dec 11, 2016 at 2:10

user557846

2 Comments

Gareth Gillman Over a year ago

I must be missing something but can't get it too work, tried every option possible, still getting the php error.....

user557846 Over a year ago

@GarethGillman you need to provide the exact code, or at least enough for us to replicate the error

Collectives™ on Stack Overflow

PHP echo array with an array?

2 Answers 2

Comments

2 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

2 Comments

Your Answer

Sign up or log in

Post as a guest

Related