Why am I getting a syntax error in setting this Array into a class?

Question

I am getting a syntax error but I don;t understand exactly why, here is my code:

class VerifyEmail {

     private $ip_pool = array("167.114.48.81","167.114.48.82"....);
     private $ip_to_use = $this->ip_pool[array_rand($this->ip_pool)]; //ERROR HERE
     .....

I tried also:

     private $ip_to_use = $ip_pool[array_rand($ip_pool)];

with no luck.

Am I missing something? Or you cannot do an array rand of a private variable when setting up the variables?

Thanks!

Try this private $ip_to_use = $this->ip_pool[array_rand($this->ip_pool, 1)]; — Parag Soni
– Parag Soni, Commented Jul 5, 2017 at 3:11
What is your PHP version? using $this when defining default values feels wrong. this logic might be more suitable in the constructor. — Scuzzy
– Scuzzy, Commented Jul 5, 2017 at 3:40

Scuzzy · Accepted Answer · 2017-07-05 03:45:14Z

1

I get the following notice in my IDE for that line

expression is not allowed as a field default value

I can only suggest moving your rand call into the __construct() method

class VerifyEmail {

  private $ip_pool = array( "167.114.48.81", "167.114.48.82");
  private $ip_to_use;

  public function __construct() {
    $this->ip_to_use = $this->ip_pool[ array_rand( $this->ip_pool ) ];
  }

}

var_dump( new VerifyEmail() );

answered Jul 5, 2017 at 3:45

Scuzzy

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Obsidian Age · Accepted Answer · 2017-07-05 03:11:36Z

1

You're currently trying to access the array as an index of the array itself.

Considering you're just trying to assign one string from within the $ip_pool array to $ip_to_use, all you need is $ip_to_use = array_rand($this->ip_pool).

class VerifyEmail {  
  private $ip_pool = array("167.114.48.81","167.114.48.82"....);
  private $ip_to_use = array_rand($this->ip_pool);
  ...

Hope this helps! :)

answered Jul 5, 2017 at 3:11

Obsidian Age

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3 Comments

Fede E. Over a year ago

then use it as $ip_to_use[0] to get the first element? BTW, private $ip_to_use = array_rand($this->ip_pool); is also marking a syntax error.

Obsidian Age Over a year ago

You can use it with just $ip_to_use. array_rand() without a second parameter only returns one random element from the array. You can specify how many elements to return with an optional secondary int parameter (so array_rand($this->ip_pool, 2) would return two random IP addresses) :)

Fede E. Over a year ago

private $ip_to_use = array_rand($this->ip_pool); is also showing a syntax error.

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