我知道有人问过类似的问题 . 但是,我找到的解决方案都无法解决我的问题 .
我正在使用Avada WordPress主题 . 我为表列表创建了一个自定义页面,要求我连接到数据库以查看元素 . 错误陈述是
致命错误:无法在第49行的/home/public_html/wp-content/themes/Avada/krtcsearcher.php中重新声明类用户
-
尽管出现错误消息,该函数仍然可以正常工作,因为它显示来自数据库的消息 .
-
使用localhost-XAMPP(没有wordpress php代码)时该文件运行良好
-
我在类上尝试过require_once或if(!class_exist)但是没有用 .
Krtcsearcher.php 错误部分看起来像那样
<?php
// Template Name: krtcsearcher
?>
<?php get_header(); ?>
<div id="content" <?php Avada()->layout->add_class( 'content_class' ); ?> <?php Avada()->layout->add_style( 'content_style' ); ?>>
<?php while ( have_posts() ) : the_post(); ?>
<?php $page_id = get_the_ID(); ?>
<div id="post-<?php the_ID(); ?>" <?php post_class(); ?>>
<?php echo avada_render_rich_snippets_for_pages(); ?>
<?php echo avada_featured_images_for_pages(); ?>
<?php
require_once '/home/public_html/wp-content/themes/Avada/config/connection.php';
if(!class_exists('DB_Connection')){
class DB_Connection {
private $connect;
function __construct() {
$this->connect = mysqli_connect(DB_HOST, DB_USER, DB_PASS, DB_NAME)
or die("Could not connect to database");
}
public function getConnection()
{
return $this->connect;
}
}
}
class Users {
private $db;
private $connection;
function __construct() {
$this -> db = new DB_Connection();
$this -> connection = $this->db->getConnection();
//echo "Database Connection Successful!";
}
function secinfo($test) {
$sql = 'Select infoc FROM informations;
$result = mysqli_query($this->connection, $sql);
//$row = $result->fetch_assoc(); //get first row of the list
//while ($row = $result->fetch_assoc()) {
while ($row = $result->fetch_row()) {
echo $row[0];
}
mysqli_close($this->connection);
}
}
?>
新图片*
一切正常,但下面显示的错误消息 .
来自Localhost-XAMPP