Struggling to select a value from another table - php

Question

Im currently constructing a database using PHP, HTML and MySQL.
I have an events table. Event_id is the primary key. When someone registers for an event they input their student_id and they also select an event from a dropbox that pulls the events from my events table. I have an event_registration table which records all entries into this form. I have event_id in my event_registration table and I'm trying to make it so that when someone selects a certain event, that event's id automatically goes into the event_registration table. My PHP code at present is this:
$student_id = $_POST['student_id'];
$title = $_POST['title'];
$sql = "select event_id from events where title LIKE $title";
$db->select_db($database);
$event_id = $db->query($sql);
$q = "INSERT INTO event_registration (";
$q .= "student_id, event_id, title";
$q .= ") VALUES (";
$q .= "'$student_id', '$event_id', '$title')";
$result = $db->query($q);
The student_id, title, and registraion_id (auto- increment) are inserting fine, but event_id is always showing up as 0, I'm not sure where I'm going wrong. Any help is appreciated.
<td style="width: 176px; height: 23px">Event Title</td>
<td style="height: 23px"><select name="title" style="width: 124px">
<?php
include ("detail.php");
$sql = mysqli_query($db, "SELECT title FROM events");
while ($row = $sql->fetch_assoc()){
?>
<option><?php echo ($row['title']); ?></option>";
<?php
}
?>
</select></td>
my dropdown code

Answer 1

The problem you seem to be having is:
$event_id = $db->query($sql);
Probably doesn't give you the number you're expecting right away, instead you need to fetch like this:
$result= $db->query($sql);
$result = $result->fetch_assoc();
$event_id = $result['event_id'];
However, based on your comment, you're constructing your dropdown with this:
<?php
include ("detail.php");
$sql = mysqli_query($db, "title FROM events");
while ($row = $sql->fetch_assoc()){
?>
<option><?php echo ($row['title']); ?></option>
<?php } ?>
You can do this instead:
<?php
include ("detail.php");
$sql = mysqli_query($db, "SELECT event_id,title FROM events");
while ($row = $sql->fetch_assoc()){
?>
<option value="<?php echo $row['event_id']; ?>"><?php echo ($row['title']); ?></option>";
<?php } ?>
Notice how I'm also adding event_id on the select caluse. WIth this the users will see the title as expected, but you'll get the id directly with POST like this:
$event_id = $_POST['event_id'];
You DO have to change the name of the select dropdown to event_id instead of title like this:
<select name="event_id">