Count With Conditional on PostgreSQL - postgresql

Question

I have a table with people and another with visits. I want to count all visits but if the person signed up with 'emp' or 'oth' on ref_signup then remove the first visit. Example:
This are my tables:
PEOPLE:
id | ref_signup
---------------------
20 | emp
30 | oth
23 | fri
VISITS
id | date
-------------------------
20 | 10-01-2019
20 | 10-05-2019
23 | 10-09-2019
23 | 10-10-2019
30 | 09-10-2019
30 | 10-07-2019
On this example the visit count should be 4 because persons with id's 20 and 30 have their ref_signup as emp or oth, so it should exclude their first visit, but count from the second and forward.
This is what I have as a query:
SELECT COUNT(*) as visit_count FROM visits
LEFT JOIN people ON people.id = visits.people_id
WHERE visits.group_id = 1
Would using a case on the count help on this case as I just want to remove one visit not all of the visits from the person.

Answer 1

Subtract from COUNT(*) the distinct number of person.ids with person.ref_signup IN ('emp', 'oth'):
SELECT
COUNT(*) -
COUNT(DISTINCT CASE WHEN p.ref_signup IN ('emp', 'oth') THEN p.id END) as visit_count
FROM visits v LEFT JOIN people p
ON p.id = v.id
See the demo.
Result:
| visit_count |
| ----------- |
| 4 |
Note: this code and demo fiddle use the column names of your sample data.

Answer 2

Premise, select the count of visits from each person, along with a synthetic column that contains a 1 if the referral was from emp or oth, a 0 otherwise. Select the sum of the count minus the sum of that column.
SELECT SUM(count) - SUM(ignore_first) FROM (SELECT COUNT(*) as count, CASE WHEN ref_signup in ('emp', 'oth') THEN 1 ELSE 0 END as ignore_first as visit_count FROM visits
LEFT JOIN people ON people.id = visits.people_id
WHERE visits.group_id = 1 GROUP BY id) a

Answer 3

where's "people_id" in your example ?
SELECT COUNT(*) as visit_count
FROM visits v
JOIN people p ON p.id = v.people_id
WHERE p.ref_signup IN ('emp','oth');
then remove the first visit.
You cannot select count and delete the first visit at same time.
DELETE FROM visits
WHERE id IN (
SELECT id
FROM visits v
JOIN people p ON p.id = v.people_id
WHERE p.ref_signup IN ('emp','oth')
ORDER BY v.id
LIMIT 1
);
edit: typos

Answer 4

First, I create the tables
create table people (id int primary key, ref_signup varchar(3));
insert into people (id, ref_signup) values (20, 'emp'), (30, 'oth'), (23, 'fri');
create table visits (people_id int not null, visit_date date not null);
insert into visits (people_id, visit_date) values (20, '10-01-2019'), (20, '10-05-2019'), (23, '10-09-2019'), (23, '10-10-2019'), (30, '09-10-2019'), (30, '10-07-2019');
You can use the row_number() window function to mark which visit is "visit number one":
select
*,
row_number() over (partition by people_id order by visit_date) as visit_num
from people
join visits
on people.id = visits.people_id
Once you have that, you can do another query on those results, and use the filter clause to count up the correct rows that match the condition where visit_num > 1 or ref_signup = 'fri':
-- wrap the first query in a WITH clause
with joined_visits as (
select
*,
row_number() over (partition by people_id order by visit_date) as visit_num
from people
join visits
on people.id = visits.people_id
)
select count(1) filter (where visit_num > 1 or ref_signup = 'fri')
from joined_visits;

Answer 5

-- First get the corrected counts for all users
WITH grouped_visits AS (
SELECT
COUNT(visits.*) -
CASE WHEN people.ref_signup IN ('emp', 'oth') THEN 1 ELSE 0 END
AS visit_count
FROM visits
INNER JOIN people ON (people.id = visits.id)
GROUP BY people.id, people.ref_signup
)
-- Then sum them
SELECT SUM(visit_count)
FROM grouped_visits;
This should give you the result you're looking for.
On a side note, I can't help but think clever use of a window function could do this in a single shot without the CTE.
EDIT: No, it can't since window functions run after needed WHERE and GROUP BY and HAVING clauses.