php - display ID in URL - php

Question

for the past few i have been trying to get the ID of the event a user registers to inside the URL in order to show all details in that event.
Ive tried Get data from MySQL database by specific id in url
But it doesnt seem to work for what i am trying to do.
Or i am doing it wrong, i got more progress with this code
<?php
if(isset($_SESSION['user_id'])){
$userid = $_SESSION['user_id'];
$QUERY3 = mysqli_query($DB, "SELECT * FROM `registration` WHERE `user_id`='$userid'");
$GETEVENTZ = mysqli_fetch_array($QUERY3);
}
?>
<?php
$eventid = $GETEVENTZ['event_id'];
echo '<li> <i class="icon icon-list"></i><span>Leaderboards</span> </li>'
?>
Just to try and get the ID in but im getting a normal page
with the url being
leaderboard.php?eventid=
and no ID after, no errors are popping up, page and everything loads, just doesnt add ID. cant seem to find the issue

Answer 1

Since your are using: $GETEVENTZ = mysqli_fetch_array($QUERY3);
Your code is expecting an array return. So you cannot just simply use:
$eventid = $GETEVENTZ['event_id'];
A quick fix can be $eventid = $GETEVENTZ[0]['event_id'] if you are expecting a single row return from your query.
But the proper way to handle this is to use mysqli_fetch_row() instead.
You can also dump the result set first to debug what your query is returning.

Answer 2

The session super global is not the one to use. Use $_GET to get information from the URL.
<?php
if(isset($GET['user_id'])){
$userid = $GET['user_id'];
$mysqli = new mysqli("localhost", "my_user", "my_password", "world");
/* check connection */
if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
$query3 = "SELECT * FROM `registration` WHERE `user_id`=?";
if($stmt = $mysqli->prepare($query3)) {
$stmt->bind_param("s", $userid);
$stmt->execut();
$result = $stmt->get_result();
}
}
?>
<?php
// this section does not demonstrate getting the value from $result
// I will update answer later, must research mysqli more.
echo '<li> <i class="icon icon-list"></i><span>Leaderboards</span> </li>'
?>
Disclaimer: I don't use mysqli, so my answer is a best guess on that portion. Be sure to research how to properly use prepared statements at http://php.net/manual/en/mysqli.prepare.php

Answer 3

Try this, Hopefully it will work.
<?php
$DB = new mysqli("localhost", "my_user", "my_password", "my_db");
/* check connection */
if ($DB->connect_errno) {
printf("Connect failed: %s\n", $DB->connect_error);
exit();
}
if(isset($_SESSION['user_id'])) {
$userid = $_SESSION['user_id'];
$QUERY3 = "SELECT * FROM registration WHERE user_id='$userid'";
$result = $DB->query($QUERY3);
$GETEVENTZ = $result->fetch_array(MYSQLI_ASSOC);
$eventid = $GETEVENTZ['event_id'];
printf ("%s", $eventid);
}
echo '<li> <i class="icon icon-list"></i><span>Leaderboards</span> </li>';
?>

Answer 4

Answer -
Session id was labeled (userid) instead of (user_id)
then after i stated if user session has user ID then print the event ID and then show page, which then it redirects me to the correct page with ID of event
<?php
if(isset($_SESSION['user_id'])){
$GETUSRID = $_SESSION['user_id'];
$GETRANKD = mysqli_query($DB, "SELECT * FROM `registrations` WHERE `user_id`='$GETUSRID'");
$GETRD = mysqli_fetch_assoc($GETRANKD);
$eventid = $GETRD['event_id'];
printf ("%s", $eventid);
}
echo '<li> <i class="icon icon-list"></i><span>Leaderboards</span> </li>';