致命錯誤：調用第15行上class-db.php中的null成員函數query（）

Question

我正在研究一個簡單的Content Management系統，並在定製一些代碼後跟隨教程，我試圖找到我做錯了什麼，但找不到問題。

反正這裏是我的SQL連接的代碼。在$result = $mysqli->query($query);

<?php 
    if (!class_exists('DB')){ 
     class DB{ 
      public function __construct(){ 
       $mysqli = new mysqli('localhost', 'root', '', 'mdatabase'); 

       if ($mysqli->connect_errno) { 
        printf('Connection Failed %s/n',$mysqli->connect_error); 
        exit(); 
       } 
       $propertyc->connection = $mysqli; 
      } 

      public function insert($query){ 
       $result = $mysqli->query($query); 

       return $result; 
      } 

      public function select(){ 


      } 
     } 
    } 

    $db = new DB; 
?> 

Warning: Creating default object from empty value in C:\wamp\www\phptesting\includes\class-db.php on line 11

Notice: Undefined variable: mysqli in C:\wamp\www\phptesting\includes\class-db.php on line 15

Fatal error: Call to a member function query() on null in C:\wamp\www\phptesting\includes\class-db.php on line 15

Answer 1

你$ mysqli的變量是不是你的類的成員發生了致命錯誤，所以它是你的數據庫的範圍::插入（）函數內部空。試試這個：

<?php 
    if (!class_exists('DB')){ 
     class DB{ 
      public function __construct(){ 
       $this->mysqli = new mysqli('localhost', 'root', '', 'mdatabase'); 

       if ($this->mysqli->connect_errno) { 
        printf('Connection Failed %s/n',$this->mysqli->connect_error); 
        exit(); 
       } 
       $propertyc->connection = $this->mysqli; 
      } 

      public function insert($query){ 
       $result = $this->mysqli->query($query); 

       return $result; 
      } 

      public function select(){ 


      } 

      protected $mysqli; 
     } 
    } 

    $db = new DB; 
?> 

此外，您的$ propertyc變量將在那裏爲空，所以$ propertyc-> connection不指向任何東西。