努力从另一个表中选择一个值

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">