Question

I have 3 InnoDB tables: emails, websites and subscriptions.

emails table has id and email columns.

websites table has id and address columns.

subscriptions table has id, email_id and website_id columns.

What I’m tring to do is supply an email and return a table with columns address and subscribed. The former is a list of all the addresses in the websites table and the latter gets value 1 if the supplied email address has an occurence in the subscriptions table with website_id set to that website, or 0 otherwise. But I’m willing to retain all the websites even if the user is not found.

The point I’m stuck is where I should change the value of the virtual column subscribed from 0 to 1 when that email has that record.

Here’s my query so far. Does anybody know how to do this?

SELECT `address`, "0" AS `subscribed`
/* 0 becomes 1 for the websites email has subscribed to */
FROM `websites` a
LEFT JOIN (
    SELECT s.`website_id` FROM `subscriptions` s
    RIGHT JOIN (
        SELECT `id` AS `email_id` FROM `emails`
        WHERE `email`='[e-mail]' LIMIT 1) e
    ON s.`email_id`=e.`email_id`) l
ON l.`website_id`=a.`id`

And here are the example outputs for the desired values for the subscribed column:

• If email is not found in the emails table all the rows get value 0
• If email is found in the emails table…
  • if it is not found in subscriptions table all the rows get value 0
  • if it is found in subscriptions table, the appropriate address rows get value 1

Let me know if I couldn’t wxplain it well. Does anytbody know what I should alter in my query?

Thanks in advance.

Answer 1

SELECT w.address, CASE WHEN s.id IS NULL THEN 0 ELSE 1 END AS subscribed
    FROM websites w
        LEFT JOIN subscriptions s
            INNER JOIN emails e
                ON s.email_id = e.id
                    AND e.email = '[e-mail]'
            ON w.id = s.website_id

You could also come up with the subscribed value this way, which is a bit more concise but also somewhat less obvious.

SELECT w.address, COALESCE(s.id/s.id, 0) AS subscribed
    FROM websites w
        LEFT JOIN subscriptions s
            INNER JOIN emails e
                ON s.email_id = e.id
                    AND e.email = '[e-mail]'
            ON w.id = s.website_id