select *
from subscription s0
where not exists (
select 1
from subscription s1
where
s0.contact_id = s1.contact_id
and s1.start_date = s0.end_date
)
order by contact_id, id
Compare start and end dates across multiple rows
-
04-06-2022 - |
Question
I have a table of subscriptions for contacts. A contact can have multiple subscriptions:
CREATE TABLE contact (
id INTEGER NOT NULL,
name TEXT,
PRIMARY KEY (id)
);
CREATE TABLE subscription (
id INTEGER NOT NULL,
contact_id INTEGER NOT NULL REFERENCES contact(id),
start_date DATE,
end_date DATE,
PRIMARY KEY (id)
);
I need to get all subscriptions for a given by contact that do not have a subscription that starts on the same date as the end date of the another subscription for the same contact.
So for the given data:
INSERT INTO contact (id, name) VALUES
(1, 'John'),
(2, 'Frank');
INSERT INTO subscription (id, contact_id, start_date, end_date) VALUES
(1, 1, '2012-01-01', '2013-01-01'),
(2, 1, '2013-01-01', '2014-01-01'),
(3, 2, '2012-01-01', '2012-09-01'),
(4, 2, '2013-01-01', '2014-01-01');
I want to get subscriptions with ids of 2, 3, 4 but not 1, because the contact 'John'
has a subscription with a start_date
on the same day (2013-01-01
) as the end_date
for subscription with id
of 1.
What is the best way to achieve this?
Solution
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