SQL PHP：如果另一个表中存在虚拟列，请选择

Question

I'm making a notifications widget on my website and I'm trying to make it so that if a notification is marked as read, that notification ID will be inserted into another table (table 'b') along with their username so that it is marked as read. Now the problem that I run into is when displaying all notifications (whether they're read or unread) I don't know how to indicate if the notification exists in the secondary table

The currently SQL query is as follows:

$qry = "SELECT * FROM notifications WHERE (notif_recipient = '$user') ORDER BY notif_date DESC";

What I'd like to do is make the query much more complex in order to indicate if a notification exists in another table, so something along the lines of:

$qry = "SELECT notif_id,notif_message,(CASE SELECT notif_is_read AS '1' WHERE notif_id.notifications = notif_id.notifications_read ELSE SELECT notif_is_read AS '0') FROM notifications WHERE (notif_recipient = '$user') ORDER BY notif_date DESC";

Is something like this possible or is it as preposterous as my lack of ability for writing SQL queries

Answer 1

As Marc B had suggested, I decided to use LEFT JOIN in order to identify which messages currently exist in the secondary table, with my query looking as such:

$qrytest = "SELECT notifications.notif_id,notifications.notif_message,notifications.notif_flag,notifications.notif_poster,notifications.notif_date,notifications_read.notif_read_count FROM notifications LEFT JOIN notifications_read ON notifications.notif_id=notifications_read.notif_id WHERE ((notif_recipient = 'all') OR (notif_recipient = '$id')) ORDER BY notif_date DESC,notif_flag DESC";

This in turn will return the results of my primary table (notifications) and if the same notification id exists in my secondary table, I decided to echo out that table's ID count, if the entry does not exist it will simply return as null

    $exists = $row['notif_read_count'];
if ($exists !== ''){
  // When NOT returning as null do something
 } else {
  // When exist returns as null do something
 }