Question

db.php中

<?php
class DB{

protected $db_name = '';
protected $db_user = '';
protected $db_pass = '';
protected $db_host = '';

public function connect() {

    $connect_db = new mysqli( $this->db_host, $this->db_user, $this->db_pass, $this->db_name );

    if ( mysqli_connect_errno() ) {

        printf("Connection failed: %s\n", mysqli_connect_error());
        exit();
    }

    return $connect_db;

}
}

在DB.php中尝试为数据库创建连接。

test.php的

<?php

require('DB.php');

$db = new DB('localhost', 'root', '', 'test');
$connection = $db->connect();


$sql ="SELECT * FROM users";
$result = $connection->query($sql);

if ($result->num_rows > 0) {
    while($row = mysqli_fetch_array($result)) {             

        array_push($responseData,
        array('uid' => $row[0],
            'u_name' => $row[1],
            'u_pass' => $row[2],
            'u_fname' => $row[3],
            'u_email' => $row[4],
            'u_status' => $row[5],
            'u_notes'=>$row[6]));
    }

        echo json_encode(array("result"=>$responseData));
    }

else {
    echo '<div class="callout callout-danger">No data found...</div>';
}

在test.php中，我试图从用户表中获取数据，但它说＆＃34;试图在第12行的test.php中获取非对象的属性。＆＃34;

请帮忙。谢谢。

Answer 1

您必须在constructor文件中创建db.php。否则你的变量都是空的。

将此添加到db.php

function __construct($hostname, $username, $pw, $db) {
    $this->db_host = $hostname;
    $this->db_user = $username;
    $this->db_pass = $pw;
    $this->db_name = $db;
}

像这样设置变量。

尝试从数据库中获取数据时尝试获取非对象的属性

1 个答案: