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在PHP / MySQL / PDO中創建配置文件信息頁面

[英]Create Profile Information Page in PHP/MySQL/PDO

嗨,我在使用PDO創建配置文件頁面時遇到問題-我可以在舊版mysql中正常工作,但是我不知道如何將數據更改為PDO-我已經在這工作了幾周了而且無所適從。 請很好,我對所有這些都很陌生。 這是起作用的代碼-舊方法

<?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'];
 }
 ?>

這就是我試圖提出的,但是我只是空白屏幕

<?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'];
}
?>

登錄在舊的MySql和PDO中都可以正常運行,但是我似乎無法在PDO中顯示用戶配置文件

//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()期望參數1為資源,對象從此行中給出。

$member = mysql_fetch_assoc($result);

我得到未定義的變量:...和PHP中的stmt致命錯誤:從此行中,在...中的非對象上調用成員函數fetch()

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

正如我之前提到的那樣,我對此很陌生,並且花了很多周的時間研究制作多個教程,並試圖弄清所有這些。 當我遇到折舊的錯誤時我只是在學習MySql,現在我切換到PDO,我知道將來我將需要從多個數據庫中為一個登錄用戶提取信息。 我沒有使用函數或類。 在此先感謝,我的狗將感謝您的幫助。

好吧,我解決了! 我將發布任何其他有此問題的代碼@Fred,謝謝!

而且我知道代碼確實有所不同,因為我設法在測試站點上獲取了它,在我的登錄過程頁面中,我更改了此代碼:

// 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");
}

在頁面中,我希望顯示配置文件,我將代碼更改為:

<?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>

我知道它仍然有些混亂,但是就像我說的那樣,我是新手

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