PHP鏈接：如何從MySQL表中獲取數據？

Question

我想在我的web開發項目中實現PHP鏈接方法。但我似乎無法做到。

class foo extends base{ 


    public $query = null; 
    public $item = array(); 

    public function __construct($connection){ 
     parent::__construct($connection); 

    } 

    public function select($query){ 
     $this->query = $query; 
     return $this; 
    } 

    public function where($query){ 
     $this->query = $query; 
     return $this; 
    } 

    public function __toString() 
    { 
     $this->item = $this->connection->fetch_assoc($this->query); 

     return var_export($this->item, true); 
    } 


} 

$connection = new db($dsn = 'mysql:host=localhost;dbname=xxx',$username = 'xxx',$password = 'xxx'); 
$foo = new foo($connection); 

$select = $foo->select("SELECT * FROM page")->where("page_id = 10 "); 
print_r($select->item); 

結果我得到的，

Array 
(
) 

但我應該得到的數據行。就像我通常這樣做，

class boo extends base{ 

    ... 

    public function select() { 

     $sql = " 
      SELECT * 
      FROM page 
      WHERE page_id = ? 
     "; 

     $item = $connection->fetch_assoc($sql,array(1)); 
     return $item; 

    } 
} 

我在鏈接方法中錯過了什麼？

編輯：

class base 
{ 
    protected $connection = null; 

    public function __construct($connection) 
    { 
     $this->connection = $connection; 

    } 


} 

，如果我只是打印$select，

print_r($select); 

結果，

foo Object 
(
    [query] => where page_id = 10 
    [item] => Array 
     (
     ) 

    [connection:protected] => db Object 
     (
      [connection] => PDO Object 
       (
       ) 

      [dsn] => mysql:host=localhost;dbname=xxx 
      [username] => xxx 
      [password] => xxx 
     ) 

) 

Answer 1

嘗試像

$select = $foo->select('*')->from('page')->where('page_id','10'); 
print_r($select); 

Answer 2

通過以下調用所有你正在做的是設置和更換foo類的$query成員變量的值：

$select = $foo->select("SELECT * FROM page")->where("page_id = 10 "); 

所以，$foo->select("SELECT * FROM page")設置$query成員變量的值，然後->where("page_id = 10 ")替換值。

另外，您沒有執行查詢。