数据库连接函数调用的折叠代码出错

Question

public function Fetch_num1($id) {
    
    echo "Hi";
    
    $sql = "SELECT * FROM editors where jid='$id'";
    
    $result = mysqli_query($connection, $sql); 
    
    $nr = mysqli_num_rows($result);
    
    return $nr;
}

它显示的警告如下：

mysqli_query()期望参数1为mysqli，空给定的mysqli_num_rows()期望参数1为mysqli_result，在

中为null

Answer 1

正如Armali在评论中所说的，您没有建立数据库连接。下面是从w3schools (https://www.w3schools.com/php/php_mysql_connect.asp)复制的典型连接代码：

 <?php
$servername = "localhost";
$username = "username";
$password = "password";

// Create connection
$conn = new mysqli($servername, $username, $password);

// Check connection
if ($conn->connect_error) {
  die("Connection failed: " . $conn->connect_error);
}
echo "Connected successfully";
?> 

将localhost、username和password替换为db名称、用户名和密码，然后将$connection替换为$conn。

另外，我认为where应该是大写字母：WHERE。