Question

我正在尝试显示用户名，电子邮件ID，但每次输出都显示为“ 1”。在我尝试获取并显示值的任何地方，它只显示1作为输出。

数据库：

CREATE TABLE `admin` (
  `id` int(11) NOT NULL,
  `username` varchar(50) NOT NULL,
  `password` varchar(255) NOT NULL,
  `created_at` datetime DEFAULT current_timestamp(),
  `email` varchar(255) DEFAULT NULL,
  `name` varchar(255) DEFAULT NULL
) ENGINE=MyISAM DEFAULT CHARSET=latin1;

PHP代码：

<?php
session_start();

if(!isset($_SESSION["loggedin"]) || $_SESSION["loggedin"] !== true){
    header("location: login.php");
    exit;
}
?>
<?php
require_once('config/config.php');
//Counting Member
$members = "SELECT count( * ) as  id FROM admin";
$member_result = $mysqli->query($members);
$total_member = mysqli_num_rows($member_result);
//Showing Admin Name
$admin_name = "SELECT `name` FROM `admin` WHERE id='1'";
$name_result = $mysqli->query($admin_name);
$name_sh = mysqli_num_rows($name_result);
//Showing Email
$admin_email = "SELECT `email` FROM `admin` WHERE id='1'";
$email_result = $mysqli->query($admin_email);
$email_sh = mysqli_num_rows($email_result);
?>

config.php

<?php
define('DB_SERVER', 'localhost');
define('DB_USERNAME', 'root');
define('DB_PASSWORD', '');
define('DB_NAME', 'test1');

$mysqli = new mysqli(DB_SERVER, DB_USERNAME, DB_PASSWORD, DB_NAME);

if($mysqli === false){
    die("ERROR: Could not connect. " . $mysqli->connect_error);
}
?>

HTML代码：

                                    <div class="account-dropdown js-dropdown">
                                        <div class="info clearfix">
                                            <div class="image">
                                                <a href="#">
                                                    <img src="images/icon/avatar-01.jpg" alt="John" />
                                                </a>
                                            </div>
                                            <div class="content">
                                                <h5 class="name">
                                                    <a href="#"><?php echo $name_sh; ?></a>
                                                </h5>
                                                <span class="email"><?php echo $email_sh; ?></span>
                                            </div>
                                        </div>

它只显示1而不是值。在此先感谢

Answer 1

那是因为您正在显示mysqli_num_rows的结果，该结果返回结果中的行数（因此始终等于1）。

使用mysqli_stmt_fetch来获取查询结果。

Answer 2

$admin_name = "SELECT `name` FROM `admin` WHERE id='1'";
$name_result = $mysqli->query($admin_name);
$result=mysqli->fetch_array($name_result,MYSQLI_ASSOC);

无法使用php获取和显示数据

2 个答案: