I'm trying to split a column from a result set into 2 columns based on the values from the column. So a user can subscribe to multiple items and the user can have 2 email addresses which can receive this subscription. The result set gives a list of subscriptions and their corresponding entries for subscribed email ids.
DB details
Table 1 - user_subscriptions
user_id
email_id - 1 for email id 1 and 2 for email id 2
subscription_id
Table 2 - subscriptions
subscription_id
subscription_name
Now I need all the subscriptions for the user whether subscribed by either of the email ids or not. So I get a result set something like this
+----------------------+----------+
| subscription_name | email_id |
+----------------------+----------+
| item1 | 1 |
| item1 | 2 |
| item2 | null |
| item3 | 1 |
| item4 | null |
| item5 | 2 |
+----------------------+----------+
So I'm looking to split the above result set into something like below
+-------------------+---------+---------+
| subscription_name | email_1 | email_2 |
+-------------------+---------+---------+
| item1 | 1 or Y | 1 or Y |
| item2 | 0 or N | 0 |
| item3 | 1 | 0 |
| item4 | 0 | 0 |
| item5 | 0 | 1 |
+-------------------+---------+---------+
Hope this question makes sense. Any help would be appreciated!
Updated -----------
Sample Data:
subscriptions - +-----------------+-------------------+ | subscription_id | subscription_name | +-----------------+-------------------+ | 1 | item1 | | 2 | item2 | | 3 | item3 | | 4 | item4 | | 5 | item5 | +-----------------+-------------------+
user_subscriptions
+---------+----------+-----------------+ | user_id | email_id | subscription_id | +---------+----------+-----------------+ | 101 | 1 | 1 | | 101 | 2 | 1 | | 101 | 1 | 3 | | 101 | 2 | 5 | | 102 | 1 | 1 | | 102 | 2 | 1 | +---------+----------+-----------------+
Expected Result:
For user_id = 101
+-----------------+-------------------+--------+--------+ | subscription_id | subscription_name | mail_1 | mail_2 | +-----------------+-------------------+--------+--------+ | 1 | item1 | Y | Y | | 2 | item2 | N | N | | 3 | item3 | Y | N | | 4 | item4 | N | N | | 5 | item5 | N | Y | +-----------------+-------------------+--------+--------+
You need a conditional aggregate:
select us.subscription_name,
-- there's at least one email
CASE WHEN MIN(us.email_id) IS NOT NULL THEN 'Y' ELSE 'N' END as email_1,
-- there's more than one email
CASE WHEN MIN(us.email_id) <> MAX(us.email_id) THEN 'Y' ELSE 'N' END as email_2
from subscriptions as s
left join user_subscriptions as us
on s.subscription_id = us.subscription_id
where us.user_id = ...
group by us.subscription_name
SELECT
S.subscription_id,
S.subscription_name,
CASE
WHEN US1.mail_ID IS NULL THEN 'N'
ELSE 'Y'
END mail_1,
CASE
WHEN US2.mail_ID IS NULL THEN 'N'
ELSE 'Y'
END mail_2
FROM subscriptions S
LEFT JOIN user_subscriptions US1
ON S.subscription_id = US1.subscription_id
AND US1.mail_id = 1
LEFT JOIN user_subscriptions US2
ON S.subscription_id = US2.subscription_id
AND US2.mail_id = 2
WHERE us1.user_id = 5 -- or use a variable @user_ID
OR us2.user_id = 5
I've not worked in sybase before, but I'm fairly sure the following SQL will translate easily (or even run directly):
SELECT
s.subscription_name,
COUNT(email_1.subscription_id) AS email_1,
COUNT(email_2.subscription_id) AS email_2
FROM subscriptions AS s
LEFT JOIN user_subscriptions AS email_1 ON (
s.subscription_id = email_1.subscription_id AND
email_1.email_id = 1
)
LEFT JOIN user_subscriptions AS email_2 ON (
s.subscription_id = email_2.subscription_id AND
email_2.email_id = 2
)
;
You could also say IF(email_1.subscription_id IS NOT NULL, 'Y', 'N')
etc in the SELECT
to return a straight-forward yes/no rather than a count etc.
It works on the principle that the list of LEFT JOIN
statements will match any "user subscription" record with email_id=1
and email_id=2
etc.
My lack of sybase knowledge disclaimer: ANSI SQL is can't perform PIVOT
- if sybase does, you could do this far more elegantly I'm sure. There's another question+answer which hints that sybase can do such things; it'd be worth your while looking there: https://stackoverflow.com/a/8114446/817132
Hope it helps!
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