Php 从表中提取值时发生数据库错误

嗨，我正在使用ajax从数据库中获取值。我与数据库的连接正常，但当我试图显示结果时，它显示了一个错误。我怎样才能解决这个问题？短暂性脑缺血发作

我的配置文件：

<?php
define('DB_HOST', 'localhost');
define('DB_NAME', 'steptwor_sscamera');
define('DB_USERNAME','root');
define('DB_PASSWORD','');
$con = mysqli_connect(DB_HOST, DB_USERNAME, DB_PASSWORD, DB_NAME);
if( mysqli_connect_error()) echo "Failed to connect to MySQL: " . mysqli_connect_error();

您错过了以下位置：

    $query = "SELECT category_name, product_name, amount FROM v_product_list WHERE UPPER($type) LIKE '".strtoupper($name)."%'";

可能的重复可能的重复
<br />
<b>Warning</b>:  mysqli_fetch_assoc() expects parameter 1 to be mysqli_result, boolean given in <b>F
:\xampp\htdocs\sscamera\ajax.php</b> on line <b>10</b><br />
[]

    $query = "SELECT category_name, product_name, amount FROM v_product_list WHERE UPPER($type) LIKE '".strtoupper($name)."%'";