php - 给出警告 : mysqli_query() expects parameter 1 to be mysqli, null。当我尝试将数据库作为 OOP 对象连接时

Question

我有一个关于将我的数据库连接到我的 php 并获取数据的问题。错误是警告:

mysqli_query() expects parameter 1 to be mysqli, null given

这是我的 2 个文件，类和实现

我的代码:

<?php

class DB{
    //this is my protected data
    protected $db_hostname = 'localhost';
    protected $db_username = 'root';
    protected $db_password = '';
    protected $db_name = 'blog';

    public function connect(){
        //connection to DB
        $connect_db = mysqli_connect($this->db_hostname, $this->db_username, $this->db_password, $this->db_name) or die(mysqli_connect_error($connect_db));
        //close the connection
        mysqli_close($connect_db);
        //for test
        echo "Greate";
    }
}
?>

第二个文件:-

$connection  = $db->connect();
$result = "SELECT * FROM blog_posts ";
$query = mysqli_query($connection, $result) or die(mysqli_error());

//loop through my data 
while($row = mysqli_fetch_array($query)){

  ?>
  <div class="well">
    //print data
    By: <?php echo $row->author ?>
    <p><?php echo $row->title ?></p>
  </div>
  <?php } ?>


?>

Answer 1

函数需要如下所示:-

public function connect(){
   //connection to DB
   $connect_db = mysqli_connect($this->db_hostname, $this->db_username, $this->db_password, $this->db_name);
   if(mysqli_connect_error()){ // you can use mysqli_connect_errorno() also
       die(mysqli_connect_error()); // remove $connect_db
   }else{
     return $connect_db; // return connection object
   }

}

也进行了如下更改:-

while($row = mysqli_fetch_assoc($query)){

  ?>
  <div class="well">
    //print data
    By: <?php echo $row['author']; ?>
    <p><?php echo $row['title']; ?></p>
  </div>
  <?php } ?>


?>

注意:- mysqli_fetch_array() 将为您提供索引+关联数组。而mysqli_fetch_assoc()只会给你关联数组，这使得你的结果数组更轻。谢谢

查看它以获得更好的错误报告:- http://php.net/manual/en/mysqli-driver.report-mode.php