当用户使用用户名登录时获取用户ID&密码 [英] Obtain User ID when a user logs in using Username & Password

问题描述

我需要输入USER表的USER_ID字段作为另一个表中的外键。
因此，我认为我需要在用户登录时启动一个会话。我很难让它在任何地方回显。我可以开始一个会话，允许我回显用户名，但不能复制它的ID。

有什么协助吗？我的处理代码如下。

处理脚本：
我也粘贴了暂时试图回显的部分进行测试。

 <？php 
 session_start（）; 
 $ error_message = array（）; 
 $ error = false; 
 $ dbhost =localhost; 
 $ dbname =xxx; 
 $ dbuser =xxx; 
 $ dbpass =xxx; 
 
 $ conn = new PDO（mysql：host = $ dbhost; dbname = $ dbname，$ dbuser，$ dbpass）; 
 
 $ user = $ _POST ['uname']; 
 $ password = $ _POST ['pword']; 
 
 if（$ user ==''）{
 $ error_message [] ='您尚未输入所需的用户名。 
 $ error = true; 
 
 if（$ password ==''）{
 $ error_message [] ='您没有输入所需的用户名'; 
 $ error = true; 
 
 
 $ result = $ conn-> prepare（SELECT * FROM USER WHERE USERNAME =：un and PASSWORD =：pw）; 
 $ result-> bindParam（'：un'，$ user）; 
 $ result-> bindParam（'：pw'，$ password）; 
 $ result-> execute（）; （$ rows = $ result-> fetch（PDO :: FETCH_NUM））{
 
 
 $ b if（$ rows> 0 ）{
 
 $ _SESSION ['Logged_In'] = true; 
 $ _SESSION ['username'] = $ user; 
 $ _SESSION ['USER_ID'] = $ rows ['USER_ID']; 
 header（'Location：index.php'）; 
} 
 
 else {
 
 $ error_message [] ='您的用户名和密码不正确。再试一次。'; 
 $ error = true; 
} 
 
 if（$ error）{
 $ _SESSION ['ERROR_MESSAGE'] = $ error_message; 
 session_write_close（）; 
 header（location：log_in.php）; 
 exit（）; 
}} 
 
？> 
  

我想要显示USER_ID：

 <？php 
 session_start（）; 
 if（isset（$ _ SESSION ['Logged_In']））
 {
 echo'< br>< br>'; 
 echo'您已登录为'; 
 echo $ _SESSION ['username']; 
 echo'< br>'; 
 echoID =。$ _ SESSION ['USER_ID']; 
 echo'< br>'; 
 echo'< a href =logout.php> 
点击此处注销。< / a>'; 
} 
 else 
 {
 echo'< br>'; 
 echo'您没有登录！< br>'; 
 echo'< a href =log_in.php>点击此处登录，< / a>< br>'; 
 echo'< a href =register.php>或点击此处注册。< / a>'; 
} 
？> 
  

解决方案

使用FETCH_ASSOC而不是FETCH_NUM

I need to input the USER_ID field of my USER table as a foreign key in another table. Therefore, I think I need to start a session when the user logs in. I'm having difficulty getting it to echo anywhere. I am able to start a session that allows me to echo the USERNAME, but cannot replicate it for the ID.

Any assistance please? My processing code is below. I have also pasted the section where I am temporarily trying to echo it for testing.

Processing script:

<?php
session_start();
$error_message = array();
$error = false;
$dbhost     = "localhost";
$dbname     = "xxx";
$dbuser     = "xxx";
$dbpass     = "xxx";

$conn = new PDO("mysql:host=$dbhost;dbname=$dbname",$dbuser,$dbpass);

$user = $_POST['uname'];
$password = $_POST['pword'];  

if($user == '') {
    $error_message[] = 'You have not entered the required username.';
    $error = true;
}
if($password == '') {
    $error_message[] = 'You have not entered the required username';
    $error = true;
}

$result = $conn->prepare("SELECT * FROM USER WHERE USERNAME= :un AND PASSWORD= :pw");
$result->bindParam(':un', $user);
$result->bindParam(':pw', $password);
$result->execute();

while ($rows = $result->fetch(PDO::FETCH_NUM)) {



if($rows > 0) {

$_SESSION['Logged_In'] = true;
$_SESSION['username'] = $user; 
$_SESSION['USER_ID'] = $rows['USER_ID'];
header('Location: index.php');
}

else{

    $error_message[] = 'Your username and password are not correct. Try again.'; 
    $error = true;
}

if($error) {
    $_SESSION['ERROR_MESSAGE'] = $error_message;
    session_write_close();
    header("location: log_in.php");
    exit();
}}

?>

Where I want the USER_ID to appear:

<?php
         session_start();
         if(isset($_SESSION['Logged_In']))
{
    echo '<br><br>';
    echo 'You are logged in as ';
    echo $_SESSION['username'];
    echo '<br>';
    echo "ID = ".$_SESSION['USER_ID'];
    echo '<br>';
    echo '<a href="logout.php">
Click here to log out.</a>';
}
else
{   
    echo '<br>';
    echo 'You are not logged in!<br>';
    echo '<a href="log_in.php">Click here to log in,</a><br>';
    echo '<a href="register.php">or click here to register.</a>';
}
?>

解决方案

use FETCH_ASSOC instead of FETCH_NUM

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