php - 无法回显数据库项目

Question

我试图在 php 中回显一些数据库项目，但似乎没有任何结果。所需的初始化 php 调用存储所有配置的 database.php，如下所示。我做错了什么？

SQL语句:

<?php
require_once("includes/initialize.php");

$userName = $_POST["name"];
$userEmail = $_POST["email"];

$sqlName = "SELECT name FROM individual";
$sqlEmail = "SELECT email FROM individual";

if ($sqlEmail == $userEmail || $sqlName == $userName){

$message = "Hi " + $userName + "this is your new password.";

echo $message;

}

?>

数据库配置位于另一个名为database.php的php文件中。

数据库.php:

<?php

require_once ("config.php");

 class MySQLDatabase {

private $connection;

function __construct() {
    $this->connection = mysqli_connect(DB_SERVER, DB_USER, DB_PASS, DB_NAME) or die
                    ("Database connection failed: " .
                    mysqli_connect_error() .
                    " (" . mysqli_connect_errno() . ")"
    );
    $db_select = mysqli_select_db($this->connection, DB_NAME);
}

public function close_connection() {
    if (isset($this->connection)) {
        mysqli_close($this->connnection);
        unset($this->connection);
    }
}

public function query($sql) {
    $result = mysqli_query($this->connection, $sql);
    $this->confirm_query($result);
    return $result;
}

private function confirm_query($result) {
    if (!$result) {
        die("Database query failed.");
    }
}

public function escape_value($string) {

    $escaped_string = mysqli_real_escape_string($this->connection, $string);
    return $escaped_string;
}

public function fetch_array($id){
    if (mysqli_fetch_array($id)) {
        return true;
    } 
    }
    }

    $database = new MySQLDatabase();
    $db = & $database;
    ?>

Answer 1

您没有运行查询.. ..或获取其结果..

$sqlEmail = "SELECT email FROM individual";

$query = $db->query($sqlEmail);

$user = $db->fetch_array($query);

var_dumpr($user);

希望这有帮助..