如何获取已登录站点的用户的会话ID

Question

I'm in need of some help with 'getting' the user id of who is logged into my website. I essentially have a users dashboard which I only want to show that particular users details. Could someone tell me if I'm on the right sort of path and what I've done wrong.

So in the dashboard page, which is where the users land after logging in, I have this at the top of the page:

$query = "SELECT * FROM `checkPointMod` WHERE `idUsers` = $_SESSION['id']";

Then in my header which is on every page I have this:

<?php
    session_start();
    require "includes/dbh.inc.php";
?>

Then in my login script I have:

$_SESSION['id'] = $row['idUsers']; //User ID
$_SESSION['uid'] = $row['uidUsers']; //Username
$_SESSION['email'] = $row['emailUsers']; //Users Email

So I thought by calling the $_SESSION['id'] at the top of my dashboard page it would just pick up who is logged in however the error I'm getting is:

syntax error, unexpected '' (T_ENCAPSED_AND_WHITESPACE), expecting '-' or identifier (T_STRING) or variable (T_VARIABLE) or number (T_NUM_STRING)

Please can someone tell me if this is not the correct way of getting which user has logged in or if it's just a simple case of syntax?

Answer 1

You can make a variable to store the id and pass it in the query to ignore this error like this

$id =  $_SESSION['id'];
$query = "SELECT * FROM `checkPointMod` WHERE `idUsers` = $id";

Or you can do this

$query = "SELECT * FROM `checkPointMod` WHERE `idUsers` = ".$_SESSION['id'].'"';

Answer 2

I think problem in your sql query.

When you use assosiative array havin key as string you should concate it with '.' operator . It can not be used in double quotes directly like simple variables.

You should use query like

$uid = $_SESSION['id'];

$query = "SELECT * FROM `checkPointMod` WHERE `idUsers` = $uid";

Or

$query = "SELECT * FROM `checkPointMod` WHERE `idUsers` = ".$_SESSION['id'];