在php类构造中使用数组将数组转换为字符串会出错

Question

Get error Notice: Array to string conversion

what's wrong in this line $this->database['dsn'] = "mysql:host=$this->database['host'];dbname=$this->database['db']";

my code

class databaseClass {

  // data variables
  private $database;


  // Construct
  public function __construct() {

    // database info
    $this->database['host'] = 'localhost';
    $this->database['db'] = 'dbname';
    $this->database['username'] = 'root';
    $this->database['password'] = '123';
    $this->database['dsn'] = "mysql:host=$this->database['host'];dbname=$this->database['db']";


  }

}

Answer 1

You need to use complex syntax (curly braces) for the variable interpolation.

"mysql:host={$this->database['host']};dbname={$this->database['db']}";

See the "Complex (curly) syntax" section in the manual on variable parsing .

Using simple syntax (without the braces) PHP is just trying to insert $this->database into the string, which gives you that notice when it converts the array to a string.

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Not directly related to that problem, but I'd suggest passing the connection info as arguments to the constructor. Hard coding them in the function body is a very inflexible way to do it. Try using a different database connection for testing, for example.

Answer 2

Your $database variable hasn't been initialised as an array before assigning values to it. Try:

private $database = [];