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构造函数应该做的不仅仅是创建一个实例吗?

oop php

0

Mona Coder · 技术社区 · 6 年前

dbcon connect() 函数连接到数据库。现在我的问题是为什么不在构造函数中连接到db?

function __construct() {
      $this->DBSERVER  = "localhost" 
      $this->DBUSERNAME = "root" 
      $this->DBPASSWORD = "" 
      $this->DBNAME    = "thedb" 

      $conn = new mysqli($this->DBSERVER, $this->DBUSERNAME, $this->DBPASSWORD, $this->DBNAME);
      if ($conn->connect_error) {
        die("Connection failed: " . $conn->connect_error);
      }     
      return $conn;
    } 
    }

<?PHP
class dbcon {
    private $DBSERVER;
    private $DBUSERNAME;
    private $DBPASSWORD;
    private $DBNAME;

    protected function connect(){
      $this->DBSERVER   = "localhost" 
      $this->DBUSERNAME = "root" 
      $this->DBPASSWORD = "" 
      $this->DBNAME     = "thedb" 

      $conn = new mysqli($this->DBSERVER, $this->DBUSERNAME, $this->DBPASSWORD, $this->DBNAME);
      if ($conn->connect_error) {
        die("Connection failed: " . $conn->connect_error);
      }     
      return $conn;
    }
}
?>

1 回复 | 直到 6 年前

1

0

Mirko Steiner 6 年前

我的第一个想法是,这可能很难进行单元测试。每次您从该类中创建实例时,必须有一个可以连接到的数据库,否则它将失败。

调用方法并在一行中创建类的实例有一个技巧:

($myDBInstance = new dbcon("localhost", "mirko", "mysecret", "mydb"))->connect();

$myDBInstance = dbcon::getInstanceAndConnect("localhost", "mirko", "mysecret", "mydb");

这可能看起来像:

public static getInstanceAndConnect(a,b,c,d) {
     $mydbcon=new dbcon(a,b,c,d);
     $mydbcon->connect();
     return $mydbcon;
}

对于懒惰的争论,我很抱歉:-)