在PHP/MySQL/PDO中创建配置文件信息页面


Create Profile Information Page in PHP/MySQL/PDO

嗨,我有一个问题,使用PDO创建一个配置文件页面-我可以得到一个工作,那是在旧mysql,但我不知道如何改变数据到PDO-我已经在这几个星期了,只是在损失。请对我好一点,我对这一切都很陌生。下面是工作的代码-旧的方式

<?php
 require_once('connection.php');
 $id=$_SESSION['SESS_MEMBER_ID'];
 $result3 = mysql_query("SELECT * FROM member where mem_id='$id'");
 while($row3 = mysql_fetch_array($result3))
 { 
 $fname=$row3['fname'];
 $lname=$row3['lname'];
 $address=$row3['address'];
 $contact=$row3['contact'];
 $picture=$row3['picture'];
 $gender=$row3['gender'];
 }
 ?>

这就是我想要的但是我只是得到一个空白的屏幕

<?php
require_once('connectpdo.php');
$id=$_SESSION['SESS_MEMBER_ID'];
$result3 = $db->prepare("SELECT * FROM member where mem_id='$id'");
$row3 = $stmt->fetch (PDO::FETCH_ASSOC);
{ 
$fname=$row3['fname'];
$lname=$row3['lname'];
$address=$row3['address'];
$contact=$row3['contact'];
$picture=$row3['picture'];
$gender=$row3['gender'];
}
?>

登录在旧MySql和PDO中都可以正常工作,但我似乎无法在PDO

中获得用户配置文件
//Create query
    $result = $conn->prepare("SELECT * FROM  member WHERE username= :xtxt  AND password= :ztzt");
    $result->bindParam(':xtxt', $username);
    $result->bindParam(':ztzt', $password);
    $result->execute();
    $rows = $result->fetch(PDO::FETCH_NUM);

    //Check whether the query was successful or not
    if($rows > 0) {
            //Login Successful
            session_regenerate_id();
            $member = mysql_fetch_assoc($result);
            $_SESSION['loggedin'] = true;
            $_SESSION['SESS_MEMBER_ID'] = $member['mem_id'];
            $_SESSION['SESS_FIRST_NAME'] = $member['firstname'];
            $_SESSION['SESS_LAST_NAME'] = $member['password'];
            $_SESSION['SESS_USERNAME'] = $member['username'];
            session_write_close();
            header("location: home.php");
            exit();

我得到PHP警告:mysql_fetch_assoc()期望参数1是resource, object given in…从这行

$member = mysql_fetch_assoc($result);

我得到未定义变量:stmt in…&PHP致命错误:在…的非对象上调用成员函数fetch()

$row3 = $stmt->fetch (PDO::FETCH_ASSOC);

正如我之前提到的,我在这方面非常新手,我花了很多周的时间研究,做了很多教程,并试图弄清楚这一切。当我得到贬值的错误,我只是学习MySql,现在我切换到PDO,我知道在未来我将需要从多个数据库为一个登录用户拉信息。我没有使用函数或类。提前感谢,我的狗会很感激你的帮助。

好了,我解决了!我将为任何有这个问题的人发布代码@Fred,谢谢!

我知道代码真的很不同,因为我设法在我的测试网站上得到它,在我的登录过程页面我改变了这个代码:

// query
$result = $conn->prepare("SELECT * FROM customers WHERE username=:hjhjhjh AND  customers_password= :asas");
$result->bindParam(':hjhjhjh', $username);
$result->bindParam(':asas', $customers_password);
$result->execute();
$rows = $result->fetch(PDO::FETCH_NUM);
if($rows > 0) {
        $_SESSION['loggedin'] = true;
        $_SESSION['username'] = $username;
        $_SESSION['customers_id']=$row['customers_id'];
        $_SESSION['SESS_CUSTOMERS_KENNEL'];//kwaon ang id sang may tyakto nga username kag password 
    header("location: http://localhost:8888/ShowMyDog/newtemplate/index.php?action=account");
}

在我想要显示个人资料的页面中,我将代码更改为:

<?Php
session_start();
if (isset($_SESSION['loggedin']) && $_SESSION['loggedin'] == true) {
    echo "Welcome to the member's area, " . $_SESSION['username'] . "!";
} else {
    echo "Please log in first to see this page.";
}
?>
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8" />
<title>Untitled Document</title>
</head>
<body>
<?Php
error_reporting(E_ERROR | E_PARSE | E_CORE_ERROR);
require "config.php"; // Database connection details. 
$customers_id=$_SESSION['loggedin']; // Collecting one record with for user
$count=$dbo->prepare("select * from customers where customers_id=:customers_id");
$count->bindParam(":customers_id",$customers_id,PDO::PARAM_INT,1);
if($count->execute()){
echo " Success <br>";
$row = $count->fetch(PDO::FETCH_ASSOC);
print_r($row);
echo "<hr>";
echo "<br>Admin id = $row[customers_id]";
echo "<br>username = $row[username]";
echo "<br>password=$row[customers_password]";
echo "<br>kennel=$row[customers_kennel]";
echo "<br>first name=$row[customers_firstname]"; 
echo "<br>last name=$row[customers_lastname]";
echo "<br>email address=$row[customers_email_address]";
echo "<br> postcode=$row[customers_postcode]";
echo "<br> street address=$row[customers_street_address]";
echo "<br> gender=$row[customers_gender]";
}
?>
</body>
</html>

我知道它仍然有点乱,但就像我说的,我是新来的