CODEWITHSUNDEEP
phpassociative-array
LordZardeck
LordZardeck

Reputation: 8293

error with using variables as associative array value in php

For some reason this is giving me the follow error: syntax error, unexpected T_VARIABLE:

 $mysql = json_decode(getenv("VCAP_SERVICES"));
 $mysql = $mysql["mysql-5.1"][0]["credentials"];

class DATABASE_CONFIG {

    public $default = array(
        'datasource' => 'Database/Mysql',
        'persistent' => false,
        'host' => 'localhost',
        'port' => $mysql['port'],  // <-- Line with error
        'login' => $mysql['username'],
        'password' => $mysql['password'],
        'database' => $mysql['name'],
        'prefix' => ''
        //'encoding' => 'utf8',
    );

    public $test = array(
        'datasource' => 'Database/Mysql',
        'persistent' => false,
        'host' => 'localhost',
        'login' => 'user',
        'password' => 'password',
        'database' => 'test_database_name',
        'prefix' => '',
        //'encoding' => 'utf8',
    );
}

I know you can use variables as values in arrays, so what is going on?

Upvotes: 2

Views: 757

Answers (3)

anon
anon

Reputation:

You're trying to reference a variable inside a class which is defined outside of that class.

PHP has no idea what $mysql is inside of that class definition.

Upvotes: 0

Mark Baker
Mark Baker

Reputation: 212452

$mysql has no scope inside the class. You need to inject it as an argument into the class constructor, and then define your array values for the class properties

Upvotes: 0

Charles
Charles

Reputation: 51411

It looks like you're trying to set the default value of a property to a variable.

You can't do that, not even inside an array. This is half PHP's parser sucking, a quarter of PHP's lack of appropriate error message, and a bit of sanity.

You'll need to do it from within the constructor instead by passing in $mysql:

$config = new DATABASE_CONFIG($mysql);

class DATABASE_CONFIG {

    public $default = array(
        'datasource' => 'Database/Mysql',
        'persistent' => false,
        'host' => 'localhost',
        'port' => null,
        'login' => null,
        'password' => null,
        'database' => null,
        'prefix' => ''
        //'encoding' => 'utf8',
    );

    public function __construct($mysql) {
        $this->default['port'] = $mysql['port']; // etc
    }

}

Upvotes: 3

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