PHP：調用一個成員函數準備（）一個非對象在

Question

我收到以下錯誤信息：Fatal error: Call to a member function prepare() on a non-object in <file location>

代碼：

$obj = new MySQLi(); 
$mysqli = $obj->connect(); 

/* Create a prepared statement */ 
if($stmt = $mysqli -> prepare(...)) { <--- error line 
    ....more code 
} 

class MySQLi { 
    public $mysqli; 

    public function connect() { 
     $this->mysqli = new mysqli('localhost', 'root', '', 'icu'); 

     if(mysqli_connect_errno()) { 
      echo "Connection Failed: " . mysqli_connect_errno(); 
      exit(); 
     } 
     mysqli_set_charset($this->mysqli, "utf8"); 
     return $this->mysqli; 
    } 

    public function con_close(){ 
     mysqli_close($con); 
    } 
} 

Answer 1

的MySQLi類沒有->connect()方法，你可能有重複的類名這是不是一個好主意來工作（這也是你得到的原因是NULL並且不能調用prepare方法）。

在庫MySQLi類，你開始在constructor連接：

$mysqli = new mysqli('localhost', 'db_user', 'db_password', 'the_database'); 

Answer 2

1. 您需要的mysqli實例：

   $mysqli = new mysqli('localhost', 'root', '', 'icu'); 
   

2. 您有空格這裏->：

   if($stmt = $mysqli -> prepare(...)).... 
   

   將其更改爲：

   if($stmt = $mysqli->prepare(...))...